JEE Advanced PYQ Work, Energy and Power – Collisions

Q 1 :

A tennis ball is dropped on a horizontal smooth surface. It bounces back to its original position after hitting the surface. The force on the ball during the collision is proportional to the length of compression of the ball. Which one of the following sketches describes the variation of its kinetic energy $K$ with time $t$ most appropriately? The figures are only illustrative and not to scale. [2014]

[IMAGE 128]
[IMAGE 129]
[IMAGE 130]
[IMAGE 131]

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(2)

As tennis ball is dropped, so initial velocity $u = 0$

$K.E. = \frac{1}{2} m v^{2} = \frac{1}{2} m {[u + a t]}^{2} = \frac{1}{2} m {[0 + g t]}^{2}$

$∴$ $K.E. = \frac{1}{2} m g^{2} t^{2}$ $∴$ $K.E. \propto t^{2} ...(i)$

i.e., The relation between $k$ and $t$ is parabolic.

First the kinetic energy will increase as per eq (i). As the ball touches the ground it starts deforming and loses its K.E., when the deformation is maximum, K.E. = 0. As the ball moves up it loses K.E. and gains gravitational potential energy in the same time interval. These characteristics are best illustrated by $k t$ graph shown in (2).

Q 2 :

A particle of mass $m$ is projected from the ground with an initial speed $u_{0}$ at an angle $α$ with the horizontal. At the highest point of its trajectory, it makes a completely inelastic collision with another identical particle, which was thrown vertically upward from the ground with the same initial speed $u_{0}$ . The angle that the composite system makes with the horizontal immediately after the collision is [2013]

$\frac{π}{4}$
$\frac{π}{4} + α$
$\frac{π}{2} - α$
$\frac{π}{2}$

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(1)

[IMAGE 132]

$Height, h = \frac{u_{0}^{2} \sin^{2} α}{2 g}$

$using v^{2} - u^{2} = 2 g h$

$v_{1}^{2} - u_{0}^{2} = 2 (- g) [\frac{u_{0}^{2} \sin^{2} α}{2 g}]$

$\Rightarrow v_{1}^{2} = u_{0}^{2} (1 - \sin^{2} α) = u_{0}^{2} \cos^{2} α$

$\Rightarrow v_{1} = u_{0} \cos α$

Applying conservation of linear momentum in Y-direction

$2 m v \sin θ = m v_{1} = m u_{0} \cos α ...(i)$

Applying conservation of linear momentum in X-direction

$2 m v \cos θ = m u_{0} \cos α ...(ii)$

$Dividing (i) and (ii) we get$

$\tan θ = 1 \Rightarrow θ = 45 ° = \frac{π}{4}$

Q 3 :

This question has statement I and statement II. Of the four choices given after the statements, choose the one that best describes the two statements. [2012]

Statement-I: A point particle of mass $m$ moving with speed $v$ collides with a stationary point particle of mass $M$ . If the maximum energy loss possible is given as $f (\frac{1}{2} m v^{2})$ , then $f = (\frac{m}{M + m}) .$

Statement-II: Maximum energy loss occurs when the particles get stuck together as a result of the collision.

Statement - I is true, Statement - II is true, Statement - II is the correct explanation of Statement - I.
Statement - I is true, Statement - II is true, Statement - II is not the correct explanation of Statement - I.
Statement - I is true, Statement - II is false.
Statement - I is false, Statement - II is true.

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(4)

Maximum energy loss $= \frac{P^{2}}{2 m} - \frac{P^{2}}{2 (m + M)}$ $[∵ K.E. = \frac{P^{2}}{2 m} = \frac{1}{2} m v^{2}]$

$= \frac{P^{2}}{2 m} [\frac{M}{(m + M)}] = \frac{1}{2} m v^{2} {\frac{M}{m + M}}$

Statement II is a case of perfectly inelastic collision.

By comparing the equation given in statement I with above equation, we get

$f = (\frac{M}{m + M}) instead of (\frac{m}{M + m})$

Hence statement I is wrong and statement II is correct.

Q 4 :

A ball of mass 0.2 kg rests on a vertical post of height 5 m. A bullet of mass 0.01 kg, traveling with a velocity V m/s in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 m and the bullet at a distance of 100 m from the foot of the post. The velocity V of the bullet is [2011]

[IMAGE 133]

$250 m/s$
$250 \sqrt{2} m/s$
$400 m/s$
$500 m/s$

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(4)

[IMAGE 134]

Let after $' t'$ time both ball and bullet hit the ground.

Then, $t = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 \times 5}{10}} = 1 sec.$

After collision let $V_{ball}$ be velocity of ball and $V_{bullet}$ be velocity of bullet.

$So, 20 = V_{ball} \times 1 \Rightarrow V_{ball} = 20 m/s$

$100 = V_{bullet} \times 1 \Rightarrow V_{bullet} = 100 m/s$

By law of conservation of momentum

$0.01 V = 0.01 \times 100 + 0.2 \times 20$

$\Rightarrow 0.01 V = 1 + 4$

$\Rightarrow V = \frac{5}{0.01} = 500 m/s$

Q 5 :

Two small particles of equal masses start moving in opposite directions from a point $A$ in a horizontal circular orbit. Their tangential velocities are $v$ and $2 v$ , respectively, as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at $A$ , these two particles will again reach the point $A$ ? [2009]

[IMAGE 135]

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(3)

According to question, between collision, the particles move with constant speed.

At first collision one particle having speed $2 v$ will rotate $2 \times 120 ° = 240 °$ while other particle having speed $v$ will rotate $120 °$ . Hence, first collision takes place at B. At first collision, they will exchange their velocities as the collision is elastic and the particles have equal masses. Again second collision takes place at C.

[IMAGE 136]

Now, as shown in figure, after two collisions they will again reach at point A.

Q 6 :

Two particles of masses $m_{1}$ and $m_{2}$ in projectile motion have velocities ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ respectively at time $t = 0$ . They collide at time $t_{0}$ . Their velocities become ${\vec{v}}_{1}'$ and ${\vec{v}}_{2}'$ at time $2 t_{0}$ while still moving in air. The value of $| (m_{1} {\vec{v}}_{1}' + m_{2} {\vec{v}}_{2}') - (m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2}) |$ is [2001]

zero
$(m_{1} + m_{2}) g t_{0}$
$\frac{1}{2} (m_{1} + m_{2}) g t_{0}$
$2 (m_{1} + m_{2}) g t_{0}$

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(4)

$F_{ext} = (m_{1} + m_{2}) g and Δ t = 2 t_{0}$

$F_{ext} = \frac{Δ p}{Δ t}$

$∴$ $Δ p = F_{ext} Δ t = (m_{1} {\vec{v}}_{1}' + m_{2} {\vec{v}}_{2}') - (m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2})$

$= (m_{1} + m_{2}) g \times 2 t_{0}$

Q 7 :

Three objects A, B and C are kept in a straight line on a frictionless horizontal surface. These have masses $m$ , $2 m$ and $m$ , respectively. The object A moves towards B with a speed 9 m/s and makes an elastic collision with it. Thereafter, B makes a completely inelastic collision with C. All motions occur on the same straight line. Find the final speed (in m/s) of the object C. [2009]

[IMAGE 137]

(4)

Hence, just after collision with A velocity of B

$v_{B} = \frac{(m_{B} - m_{A}) u_{B}}{m_{B} + m_{A}} + \frac{2 m_{A} u_{A}}{m_{A} + m_{B}} = \frac{0 + 2 m \times 9}{m + 2 m} = 6 m/s$

[IMAGE 138]

The collision between B and C is completely inelastic

$∴$ $m_{B} v_{B} = (m_{B} + m_{C}) v$

$\Rightarrow v = \frac{6 \times 2 m}{2 m + m} = 4 m/s .$

Q 8 :

A bob of mass $m$ , suspended by a string of length $l_{1}$ , is given a minimum velocity required to complete a full circle in the vertical plane. At the highest point, it collides elastically with another bob of mass $m$ suspended by a string of length $l_{2}$ , which is initially at rest. Both the strings are massless and inextensible. If the second bob, after collision, acquires the minimum speed required to complete a full circle in the vertical plane, the ratio $\frac{l_{1}}{l_{2}}$ is [2013]

(5)

Due to elastic head-on collision of equal mass $m$ of bob, velocity at the highest point of bob tied to string $ℓ_{1}$ is acquired by the bob tied to string $ℓ_{2}$ .

$∴$ $\sqrt{g ℓ_{1}} = \sqrt{5 g ℓ_{2}} \Rightarrow \frac{ℓ_{1}}{ℓ_{2}} = 5$

Q 9 :

A ball is projected from the ground at an angle of $45 °$ with the horizontal surface. It reaches a maximum height of $120 m$ and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of $30 °$ with the horizontal surface. The maximum height it reaches after the bounce, in metres, is ______ . [2018]

(30)

Maximum height,

$H = \frac{u^{2} \sin^{2} θ}{2 g} \Rightarrow 120 = \frac{u^{2} (\frac{1}{2})}{2 g}$

$∴$ $u^{2} = 480 g$

Upon hitting the ground, it loses half of its kinetic energy

$∴$ ${K.E}_{initial} = \frac{1}{2} m u^{2} = 240 m g$

${K.E}_{final} = \frac{1}{2} (240 m g) = 120 m g$

$∴$ $\frac{1}{2} m v^{2} = 120 m g \Rightarrow v^{2} = 240 g$

After the bounce, the maximum height the ball reaches

$∴$ $H' = \frac{v^{2} \sin^{2} θ}{2 g} = \frac{240 g \times (\frac{1}{4})}{2 g} = 30 m$

Q 10 :

A slide with a frictionless curved surface, which becomes horizontal at its lower end, is fixed on the terrace of a building of height 3h from the ground, as shown in the figure. A spherical ball of mass $m$ is released on the slide from rest at a height $h$ from the top of the terrace. The ball leaves the slide with a velocity ${\vec{u}}_{0} = u_{0} \hat{x}$ and falls on the ground at a distance $d$ from the building making an angle $θ$ with the horizontal. It bounces off with a velocity $\vec{v}$ and reaches a maximum height $h_{1}$ . The acceleration due to gravity is $g$ and the coefficient of restitution of the ground is $1 / \sqrt{3}$ . Which of the following statement(s) is(are) correct? [2023]

[IMAGE 139]

${\vec{u}}_{0} = \sqrt{2 g h} \hat{x}$
$\vec{v} = \sqrt{2 g h} (\hat{x} - \hat{z})$
$θ = 60 °$
$\frac{d}{h_{1}} = 2 \sqrt{3}$

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Select one or more options

(1, 3, 4)

[IMAGE 140]

$From v^{2} - u^{2} = 2 g h$

$0^{2} - u_{0}^{2} = 2 g h$ $∴ u_{0} = \sqrt{2 g h}$ $∴ {\vec{u}}_{0} = \sqrt{2 g h} \hat{x}$

$and v_{z} = \sqrt{2 g (3 h)}$

$∴$ $\tan θ = \frac{v_{z}}{u_{0}} = \frac{\sqrt{2 g (3 h)}}{\sqrt{2 g h}} = \sqrt{3}$ $∴$ $θ = 60 °$

$Distance d = u_{0} t = \sqrt{2 g h} \times \sqrt{\frac{2 \times 3 h}{g}} = 2 \sqrt{3} h$

After collision only velocity along z-direction will change

$v_{1} = e v_{z} = \frac{1}{\sqrt{3}} \times \sqrt{2 g (3 h)} = \sqrt{2 g h}$

$∴$ $\vec{v} = v_{1} \hat{k} + u_{0} \hat{i} = \sqrt{2 g h} \hat{k} + \sqrt{2 g h} \hat{i} = \sqrt{2 g h} [\hat{i} + \hat{k}]$

$Height h_{1} = \frac{v_{1}^{2}}{2 g} = \frac{{(\sqrt{2 g h})}^{2}}{2 g} = h$

$∴$ $\frac{d}{h_{1}} = \frac{2 \sqrt{3} h}{h} = 2 \sqrt{3}$

Therefore options (1, 3, 4) are correct.

Q 11 :

A small particle of mass $m$ moving inside a heavy, hollow and straight tube along the tube axis undergoes elastic collision at two ends. The tube has no friction and it is closed at one end by a flat surface while the other end is fitted with a heavy movable piston as shown in figure. When the distance of the piston from closed end is $L = L_{0}$ the particle speed is $v = v_{0}$ . The piston is moved inward at a very low speed $V$ such that $V ≪ \frac{d L}{L} v_{0}$ , where $d L$ is the infinitesimal displacement of the piston. Which of the following statement(s) is/are correct? [2019]

[IMAGE 141]

The particle's kinetic energy increases by a factor of 4 when the piston is moved inward from $L_{0}$ to $\frac{1}{2} L_{0}$
If the piston moves inward by $d L$ , the particle speed increases by $2 v \frac{d L}{L}$
The rate at which the particle strikes the piston is $\frac{v}{L}$
After each collision with the piston, the particle speed increases by $2 V$

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Select one or more options

(1, 4)

[IMAGE 142]

When the small particle moving with velocity $v_{0}$ undergoes an elastic collision with the heavy movable piston moving with velocity $v$ , it acquires a new velocity $v_{0} + 2 v$ . So, the increase in velocity after every collision is $2 v$ .

Time period of collision when the piston is at a distance $' L'$ from the closed end

$T = \frac{distance}{speed} = \frac{2 L}{v'}$

Where $v'$ is the speed of the particle at that time.

$∴$ Frequency or rate at which the particle strikes the piston $= \frac{v'}{2 L}$

The rate of change of speed of the particle

$= \frac{d v'}{d t} = (frequency) \times 2 v$ $∴$ $d v' = \frac{v'}{2 L} 2 v d t$

$∴$ $\frac{d v'}{v'} = \frac{v d t}{L} = \frac{- d L}{L}$

Where $d L$ is the distance travelled by the piston in time dt. The minus sign indicates decrease in $' L'$ with time.

$∴$ $\int_{V_{0}}^{V} \frac{d v'}{v'} = - \int_{L_{0}}^{x} \frac{d L}{L}$

$∴$ $l n (\frac{v'}{v_{0}}) = - l n (\frac{L}{L_{0}}) or | v' | = \frac{v_{0} L_{0}}{L}$

$When L = \frac{L_{0}}{2}, we have | v' | = \frac{v_{0} L_{0}}{L_{0} / 2} = 2 V_{0}$

$∴$ ${K.E}_{L_{0} / 2} = \frac{1}{2} m {(2 v_{0})}^{2}$

$∴$ $K . E_{L_{0}} = \frac{1}{2} m v_{0}^{2}$

$∴$ $\frac{{K.E}_{L_{0} / 2}}{{K.E}_{L_{0}}} = 4$

Q 12 :

A flat plate is moving normal to its plane through a gas under the action of a constant force F. The gas is kept at a very low pressure. The speed of the plate v is much less than the average speed u of the gas molecules. Which of the following options is/are true? [2017]

The pressure difference between the leading and trailing faces of the plate is proportional to uv
The resistive force experienced by the plate is proportional to v
The plate will continue to move with constant non-zero acceleration, at all times
At a later time the external force F balances the resistive force

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Select one or more options

(1, 2, 4)

[IMAGE 143]

$1 = \frac{v_{1} - v}{v + u}$ $1 = \frac{v + v_{2}}{u - v}$

$∴$ $v_{1} = u + 2 v$ $∴$ $v_{2} = u - 2 v$

$∴$ $Δ v_{1} = 2 u + 2 v$ and $Δ v_{2} = 2 u - 2 v$

$Now F_{1} = \frac{d p_{1}}{d t} = ρ A (u + v) (2 u + 2 v)$

$and F_{2} = \frac{d p_{2}}{d t} = ρ A (u - v) (2 u - 2 v)$

$∴$ $F_{1} = 2 ρ A {(u + v)}^{2} and F_{2} = 2 ρ A {(u - v)}^{2}$

$Δ F is the net force due to the air molecules on the plate.$

[IMAGE 144]

$Δ F = F_{1} - F_{2} = 8 ρ A u v$ $∴$ $P = \frac{Δ F}{A} = 8 ρ u v$

$The net force F_{net} = F - Δ F = m a$

$∴$ $F - 8 ρ A u v = m a$

Due to viscosity, the plate will eventually reach terminal velocity. So now the plate will move with constant velocity.

Q 13 :

A point mass of 1 kg collides elastically with a stationary point mass of 5 kg. After their collision, the 1 kg mass reverses its direction and moves with a speed of $2 {ms}^{- 1}$ . Which of the following statement(s) is (are) correct for the system of these two masses? [2010]

Total momentum of the system is $3 {kg ms}^{- 1}$
Momentum of 5 kg mass after collision is $4 {kg ms}^{- 1}$
Kinetic energy of the centre of mass is 0.75 J
Total kinetic energy of the system is 4J

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Select one or more options

(1, 3)

According to law of conservation of linear momentum

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$

$1 \times u_{1} + 5 \times 0 = 1 (- 2) + 5 (v_{2})$

$\Rightarrow u_{1} = - 2 + 5 v_{2}$ ...(i)

The coefficient of restitution

$e = \frac{v_{2} - v_{1}}{u_{1} - u_{2}} = 1$

$\Rightarrow 1 = \frac{v_{2} - (- 2)}{u_{1} - 0} \Rightarrow u_{1} = v_{2} + 2$ ...(ii)

$From eq (i) & (ii), u_{1} = 3 m/s and v_{2} = 1 m/s$

Hence total momentum of the system = 3 kg m/s and ${K.E}_{cm} = 0.75 J$

Q 14 :

The balls, having linear momenta ${\vec{p}}_{1} = \vec{p} i$ and ${\vec{p}}_{2} = - \vec{p} i$ , undergo a collision in free space. There is no external force acting on the balls. Let ${\vec{p}}_{1}'$ and ${\vec{p}}_{2}'$ be their final momenta. The following option(s) is (are) NOT ALLOWED for any non-zero value of $p$ , $a_{1}$ , $a_{2}$ , $b_{1}$ , $b_{2}$ , $c_{1}$ and $c_{2}$ . [2008]

${\vec{p}}_{1}' = a_{1} \hat{i} + b_{1} \hat{j} + c_{1} \hat{k}$
${\vec{p}}_{2}' = a_{2} \hat{i} + b_{2} \hat{j}$
${\vec{p}}_{1}' = c_{1} \hat{k}$
${\vec{p}}_{2}' = c_{2} \hat{k}$
${\vec{p}}_{1}' = a_{1} \hat{i} + b_{1} \hat{j} + c_{1} \hat{k}$
${\vec{p}}_{2}' = a_{2} \hat{i} + b_{2} \hat{j} - c_{1} \hat{k}$
${\vec{p}}_{1}' = a_{1} \hat{i} + b_{1} \hat{j}$
${\vec{p}}_{2}' = a_{2} \hat{i} + b_{1} \hat{j}$

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Select one or more options

(1, 4)

From law of conservation of linear momentum

The initial linear momentum of the system, $p \hat{i} - p \hat{i} = 0$

$∴$ Final linear momentum should also be zero i.e., $p_{1}' + p_{2}' = 0$

$Option 1:$

$p_{1}' + p_{2}' = (a_{1} + a_{2}) \hat{i} + (b_{1} + b_{2}) \hat{j} + c_{1} \hat{k} = Final momentum$

It is given that $a_{1}, b_{1}, c_{1}, a_{2}, b_{2}$ and $c_{2}$ have non-zero values.

If $a_{1} = x$ and $a_{2} = - x,$ also if $b_{1} = y$ and $b_{2} = - y$ , then the $\hat{i}$ and $\hat{j}$ components become zero. But the third term having $\hat{k}$ component is non-zero. This gives a definite final momentum to the system which violates conservation of linear momentum.

$∴$ Option 1 is wrong.

$Option 4:$

$p_{1}' + p_{2}' = (a_{1} + a_{2}) \hat{i} + 2 b_{1} \hat{j} \neq 0 because b_{1} \neq 0$

Following the same reasoning as above the option 4 is also wrong.

Q 15 :

A small block of mass M moves on a frictionless surface of an inclined plane, as shown in figure. The angle of the incline suddenly changes from $60 °$ to $30 °$ at point B. The block is initially at rest at A. Assume that collisions between the block and the incline are totally inelastic ( $g = 10 {m/s}^{2}$ ). [2008]

[IMAGE 145]

Q. The speed of the block at point B immediately after it strikes the second incline is --

[IMAGE 146]

$\sqrt{60} m/s$
$\sqrt{45} m/s$
$\sqrt{30} m/s$
$\sqrt{15} m/s$

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(2)

[IMAGE 147]

As the inclined plane is frictionless,

$The K.E. at B = P.E. at A$

$\frac{1}{2} m v^{2} = m g h \Rightarrow v = \sqrt{2 g h}$

$In ∆ A D B, \tan 60 ° = \frac{h}{\sqrt{3}} \Rightarrow h = 3 m$

$∴$ $v = \sqrt{6 g} = \sqrt{60} m/s$

This is the velocity of the block just before collision. This velocity makes an angle of $30 °$ with the vertical. Also in right-angled triangle $B E C, ∠ E B C = 60 °$ . Therefore $v$ makes an angle of $30 °$ with the second inclined plane BC. The component of $v$ along BC is $v \cos 30 °$ .

It is given that the collision at B is perfectly inelastic, therefore the impact forces act normal to the plane such that the vertical component of velocity becomes zero. The component of velocity along the incline BC remains unchanged and is equal to $v \cos 30 °$

$= \sqrt{60} \times \frac{\sqrt{3}}{2} = \sqrt{\frac{180}{4}} = \sqrt{45} m/s$

Q 16 :

A small block of mass M moves on a frictionless surface of an inclined plane, as shown in figure. The angle of the incline suddenly changes from $60 °$ to $30 °$ at point B. The block is initially at rest at A. Assume that collisions between the block and the incline are totally inelastic ( $g = 10 {m/s}^{2}$ ). [2008]

[IMAGE 148]

Q. The speed of the block at point C, immediately before it leaves the second incline is

$\sqrt{120} m/s$
$\sqrt{105} m/s$
$\sqrt{90} m/s$
$\sqrt{75} m/s$

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(2)

$In ∆ B C E, \tan 30 ° = \frac{B E}{C E} \Rightarrow \frac{1}{\sqrt{3}} = \frac{B E}{3 \sqrt{3}} \Rightarrow B E = 3 m$

From mechanical energy conservation principle,

Mechanical energy at B = mechanical energy at C

$\frac{1}{2} M {(\sqrt{45})}^{2} + M \times 10 \times 3 = \frac{1}{2} M v_{c}^{2}$

$45 + 60 = v_{c}^{2}$ $∴$ $v_{c} = \sqrt{105} m/s$

Q 17 :

A small block of mass M moves on a frictionless surface of an inclined plane, as shown in figure. The angle of the incline suddenly changes from $60 °$ to $30 °$ at point B. The block is initially at rest at A. Assume that collisions between the block and the incline are totally inelastic ( $g = 10 {m/s}^{2}$ ). [2008]

[IMAGE 149]

Q. If collision between the block and the incline is completely elastic, then the vertical (upward) component of the velocity of the block at point B, immediately after it strikes the second incline is --

$\sqrt{30} m/s$
$\sqrt{15} m/s$
$0$
$- \sqrt{15} m/s$

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(3)

[IMAGE 150]

The velocity of the block along BC just before collision is $v \cos 30 °$ . The impact forces act perpendicular to the surface so the component of velocity along the incline remains unchanged.

Also, since the collision is elastic, the vertical component of velocity $(v \sin 30 °)$ before collision changes in direction, the magnitude remaining the same as shown in the figure. So the rectangular components of velocity after collision are as shown in the figure. This means that the final velocity of the block should be horizontal making an angle $30 °$ with BC. Therefore the vertical component of the final velocity of the block is zero.

Q 18 :

STATEMENT–1: In an elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before the collision.

STATEMENT–2: In an elastic collision, the linear momentum of the system is conserved. [2007]

Statement–1 is True, Statement–2 is True; Statement–2 is a correct explanation for Statement–1.
Statement–1 is True, Statement–2 is True; Statement–2 is NOT a correct explanation for Statement–1.
Statement–1 is True, Statement–2 is False.
Statement–1 is False, Statement–2 is True.

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(4)

Statement 1: For an elastic collision, the coefficient of restitution = 1

$e = \frac{| v_{2} - v_{1} |}{| u_{1} - u_{2} |} \Rightarrow$ $| v_{2} - v_{1} |$ $=$ $| u_{1} - u_{2} |$

$\Rightarrow$ Relative velocity after collision is equal to relative velocity before collision. But in the statement relative speed is given.

Statement 2: Linear momentum remains conserved in an elastic collision. This statement is true.

Topic Question Set

Q 12 :

A flat plate is moving normal to its plane through a gas under the action of a constant force F. The gas is kept at a very low pressure. The speed of the plate v is much less than the average speed u of the gas molecules. Which of the following options is/are true? [2017]

Q 13 :

A point mass of 1 kg collides elastically with a stationary point mass of 5 kg. After their collision, the 1 kg mass reverses its direction and moves with a speed of $2 {ms}^{- 1}$ . Which of the following statement(s) is (are) correct for the system of these two masses? [2010]

Q 18 :

STATEMENT–1: In an elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before the collision.

STATEMENT–2: In an elastic collision, the linear momentum of the system is conserved. [2007]

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Topic Question Set

Q 6 : Two particles of masses m1 and m2 in projectile motion have velocities v→1 and v→2 respectively at time t=0. They collide at time t0. Their velocities become v→1' and v→2' at time 2t0 while still moving in air. The value of |(m1v→1'+m2v→2')-(m1v→1+m2v→2)| is [2001]

Q 12 : A flat plate is moving normal to its plane through a gas under the action of a constant force F. The gas is kept at a very low pressure. The speed of the plate v is much less than the average speed u of the gas molecules. Which of the following options is/are true? [2017]

Q 13 : A point mass of 1 kg collides elastically with a stationary point mass of 5 kg. After their collision, the 1 kg mass reverses its direction and moves with a speed of 2 ms-1. Which of the following statement(s) is (are) correct for the system of these two masses? [2010]

Q 18 : STATEMENT–1: In an elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before the collision. STATEMENT–2: In an elastic collision, the linear momentum of the system is conserved. [2007]

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Q 12 :

A flat plate is moving normal to its plane through a gas under the action of a constant force F. The gas is kept at a very low pressure. The speed of the plate v is much less than the average speed u of the gas molecules. Which of the following options is/are true? [2017]

Q 13 :

A point mass of 1 kg collides elastically with a stationary point mass of 5 kg. After their collision, the 1 kg mass reverses its direction and moves with a speed of $2 {ms}^{- 1}$ . Which of the following statement(s) is (are) correct for the system of these two masses? [2010]

Q 18 :

STATEMENT–1: In an elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before the collision.

STATEMENT–2: In an elastic collision, the linear momentum of the system is conserved. [2007]