Work, Energy and Power NEET PYQ | Important Previous Year Questions

Q 1 :
Two bodies A and B of same mass undergo completely inelastic one dimensional collision. The body A moves with velocity $v_{1}$ while body B is at rest before collision. The velocity of the system after collision is $v_{2}$ . The ratio $v_{1} : v_{2}$ is [2024]

1 : 2
2 : 1
4 : 1
1 : 4

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2

3

4

(2)

Given, $m_{A} = m_{B} = m,$ $u_{A} = v_{1},$ $u_{B} = 0$

In completely inelastic collision when two bodies have same mass and one is at rest, then after collision they stick together and move with common velocity $v_{2}$ .

By conservation of momentum,

$m_{A} u_{A} + m_{B} u_{B} = (m_{A} + m_{B}) v_{2}$

$m v_{1} + 0 = 2 m v_{2}$

$∴$ $\frac{v_{1}}{v_{2}} = 2$

Q 2 :
A bullet of mass $m$ hits a block of mass $M$ elastically. The transfer of energy is the maximum, when: [2023]

$M = m$
$M = 2 m$
$M ≪ m$
$M ≫ m$

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2

3

4

(1)

In an elastic collision kinetic energy is conserved, therefore maximum energy is transferred when M = m because when M = m, velocities will get exchanged. The bullet will come to rest and block will start moving with the speed of bullet.

Q 3 :
Body A of mass $4 m$ moving with speed $u$ collides with another body B of mass $2 m$ , at rest. The collision is head-on and elastic in nature. After the collision, the fraction of energy lost by the colliding body A is [2019]

5/9
1/9
8/9
4/9

1

2

3

4

(3)

According to conservation of momentum,

$4 m u_{1} = 4 m v_{1} + 2 m v_{2} \Rightarrow 2 (u_{1} - v_{1}) = v_{2}$ ...(i)

From conservation of energy,

$\frac{1}{2} (4 m) u_{1}^{2} = \frac{1}{2} (4 m) v_{1}^{2} + \frac{1}{2} (2 m) v_{2}^{2}$

$\Rightarrow 2 (u_{1}^{2} - v_{1}^{2}) = v_{2}^{2}$ ...(ii)

From (i) and (ii), $2 (u_{1}^{2} - v_{1}^{2}) = 4 {(u_{1} - v_{1})}^{2}$

$3 v_{1} = u_{1}$ ...(iii)

Now, fraction of loss in kinetic energy for mass $4 m,$

$\frac{Δ K}{K_{i}} = \frac{K_{i} - K_{f}}{K_{i}} = \frac{\frac{1}{2} (4 m) u_{1}^{2} - \frac{1}{2} (4 m) v_{1}^{2}}{\frac{1}{2} (4 m) u_{1}^{2}}$ ...(iv)

Substituting (iii) in (iv), we get $\frac{Δ K}{K_{i}} = \frac{8}{9}$

Q 4 :
A moving block having mass $m$ , collides with another stationary block having mass $4 m$ . The lighter block comes to rest after collision. When the initial velocity of the lighter block is $v$ , then the value of coefficient of restitution (e) will be [2018]

0.5
0.25
0.8
0.4

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2

3

4

(2)

Let final velocity of the block of mass, $4 m = v'$

Initial velocity of block of mass $4 m = 0$

Final velocity of block of mass $m = 0$

According to law of conservation of linear momentum,

$m v + 4 m \times 0 = 4 m v' + 0$ $\Rightarrow v' = \frac{v}{4}$

Coefficient of restitution,

$e = \frac{Relative velocity of separation}{Relative velocity of approach}$ $= \frac{v / 4}{v} = 0.25$

Q 5 :
A bullet of mass 10 g moving horizontally with a velocity of $400$ ${ms}^{- 1}$ strikes a wooden block of mass 2 kg which is suspended by a light inextensible string of length 5 m. As a result, the centre of gravity of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be [2016]

$100$ ${ms}^{- 1}$
$80$ ${ms}^{- 1}$
$120$ ${ms}^{- 1}$
$160$ ${ms}^{- 1}$

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2

3

4

(3)

Mass of bullet, $m = 10$ $g = 0.01$ $kg$

Initial speed of bullet, $u = 400$ ${ms}^{- 1}$

Mass of block, $M = 2$ $kg$

Length of string, $l = 5$ $m$

Speed of the block after collision $= v_{1}$

Speed of the bullet on emerging from block, $v = ?$

[IMAGE 51]

Using energy conservation principle for the block,

${(KE + PE)}_{Reference} = {(KE + PE)}_{h}$

$\Rightarrow \frac{1}{2} M v_{1}^{2} = M g h or, v_{1} = \sqrt{2 g h}$

$v_{1} = \sqrt{2 \times 10 \times 0.1} = \sqrt{2}$ ${ms}^{- 1}$

Using momentum conservation principle for block and bullet system,

${(M \times 0 + m u)}_{Before collision} = {(M \times v_{1} + m v)}_{After collision}$

$\Rightarrow 0.01 \times 400 = 2 \sqrt{2} + 0.01 \times v$

$\Rightarrow v = \frac{4 - 2 \sqrt{2}}{0.01} = 117.15 {ms}^{- 1} \approx 120 {ms}^{- 1}$

Q 6 :
Two identical balls A and B having velocities of $0.5$ ${ms}^{- 1}$ and $- 0.3 {ms}^{- 1}$ respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will be [2016, 1994, 1991]

$- 0.5 {ms}^{- 1}$ and $0.3 {ms}^{- 1}$
$0.5 {ms}^{- 1}$ and $- 0.3 {ms}^{- 1}$
$- 0.3 {ms}^{- 1}$ and $0.5 {ms}^{- 1}$
$0.3 {ms}^{- 1}$ and $0.5 {ms}^{- 1}$

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2

3

4

(2)

Masses of the balls are same and collision is elastic, so their velocity will be interchanged after collision.

Q 7 :
Two particles A and B, move with constant velocities ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ . At the initial moment their position vectors are ${\vec{r}}_{1}$ and ${\vec{r}}_{2}$ respectively. The condition for particles A and B for their collision is [2015]

${\vec{r}}_{1} \times {\vec{v}}_{1} = {\vec{r}}_{2} \times {\vec{v}}_{2}$
${\vec{r}}_{1} - {\vec{r}}_{2} = {\vec{v}}_{1} - {\vec{v}}_{2}$
$\frac{{\vec{r}}_{1} - {\vec{r}}_{2}}{| {\vec{r}}_{1} - {\vec{r}}_{2} |} = \frac{{\vec{v}}_{2} - {\vec{v}}_{1}}{| {\vec{v}}_{2} - {\vec{v}}_{1} |}$
${\vec{r}}_{1} \cdot {\vec{v}}_{1} = {\vec{r}}_{2} \cdot {\vec{v}}_{2}$

1

2

3

4

(3)

Let the particles A and B collide at time $t$ . For their collision, the position vectors of both particles should be the same at time $t$ , i.e.,

${\vec{r}}_{1} + {\vec{v}}_{1} t = {\vec{r}}_{2} + {\vec{v}}_{2} t$ $;$ ${\vec{r}}_{1} - {\vec{r}}_{2} = {\vec{v}}_{2} t - {\vec{v}}_{1} t = ({\vec{v}}_{2} - {\vec{v}}_{1}) t$ ...(i)

Also, $| {\vec{r}}_{1} - {\vec{r}}_{2} |$ $=$ $| {\vec{v}}_{2} - {\vec{v}}_{1} |$ $t or t = \frac{| {\vec{r}}_{1} - {\vec{r}}_{2} |}{| {\vec{v}}_{2} - {\vec{v}}_{1} |}$

Substituting this value of $t$ in equation (i), we get

${\vec{r}}_{1} - {\vec{r}}_{2} = ({\vec{v}}_{2} - {\vec{v}}_{1}) \frac{| {\vec{r}}_{1} - {\vec{r}}_{2} |}{| {\vec{v}}_{2} - {\vec{v}}_{1} |}$ or $\frac{{\vec{r}}_{1} - {\vec{r}}_{2}}{| {\vec{r}}_{1} - {\vec{r}}_{2} |} = \frac{({\vec{v}}_{2} - {\vec{v}}_{1})}{| {\vec{v}}_{2} - {\vec{v}}_{1} |}$

Q 8 :
A ball is thrown vertically downwards from a height of 20 m with an initial velocity $v_{0}$ . It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity $v_{0}$ is (Take $g = 10 {ms}^{- 2}$ ) [2015]

28 ${ms}^{- 1}$
10 ${ms}^{- 1}$
14 ${ms}^{- 1}$
20 ${ms}^{- 1}$

1

2

3

4

(4)

[IMAGE 52]

The situation is shown in the figure.

Let $v$ be the velocity of the ball with which it collides with the ground. Then according to the law of conservation of energy,

Gain in kinetic energy = Loss in potential energy

i.e. $\frac{1}{2} m v^{2} - \frac{1}{2} m v_{0}^{2} = m g h$ (where $m$ is the mass of the ball)

or $v^{2} - v_{0}^{2} = 2 g h$ ...(i)

Now, when the ball collides with the ground, 50% of its energy is lost and it rebounds to the same height $h$ .

$∴$ $\frac{50}{100} (\frac{1}{2} m v^{2}) = m g h$

$\frac{1}{4} v^{2} = g h or v^{2} = 4 g h$

Substituting this value of $v^{2}$ in equation (i), we get

$4 g h - v_{0}^{2} = 2 g h$

or $v_{0}^{2} = 4 g h - 2 g h = 2 g h or v_{0} = \sqrt{2 g h}$

Here, $g = 10 {ms}^{- 2}$ and $h = 20 m$

$∴$ $v_{0} = \sqrt{2 (10 {ms}^{- 2}) (20 m)} = 20 {ms}^{- 1}$

Q 9 :
On a frictionless surface, a block of mass $M$ moving at speed $v$ collides elastically with another block of same mass $M$ which is initially at rest. After collision the first block moves at an angle $θ$ to its initial direction and has a speed $v / 3$ . The second block's speed after the collision is [2015]

$\frac{3}{\sqrt{2}} v$
$\frac{\sqrt{3}}{2} v$
$\frac{2 \sqrt{2}}{3} v$
$\frac{3}{4} v$

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(3)

The situation is shown in the figure.

[IMAGE 53]

Let $v'$ be speed of second block after the collision.

As the collision is elastic, so kinetic energy is conserved.

According to conservation of kinetic energy,

$\frac{1}{2} M v^{2} + 0 = \frac{1}{2} M {(\frac{v}{3})}^{2} + \frac{1}{2} M v'^{2}$

$v^{2} = \frac{v^{2}}{9} + v'^{2} or v'^{2} = v^{2} - \frac{v^{2}}{9} = \frac{9 v^{2} - v^{2}}{9} = \frac{8}{9} v^{2}$

$v' = \sqrt{\frac{8}{9} v^{2}} = \frac{\sqrt{8}}{3} v = \frac{2 \sqrt{2}}{3} v$

Q 10 :
Two particles of masses $m_{1}, m_{2}$ move with initial velocities $u_{1}$ and $u_{2}$ . On collision, one of the particles gets excited to a higher level, after absorbing energy $ε$ . If final velocities of particles be $v_{1}$ and $v_{2}$ , then we must have [2015]

$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} - ε = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$
$\frac{1}{2} m_{1}^{2} u_{1}^{2} + \frac{1}{2} m_{2}^{2} u_{2}^{2} + ε = \frac{1}{2} m_{1}^{2} v_{1}^{2} + \frac{1}{2} m_{2}^{2} v_{2}^{2}$
$m_{1}^{2} u_{1} + m_{2}^{2} u_{2} - ε = m_{1}^{2} v_{1} + m_{2}^{2} v_{2}$
$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2} - ε$

1

2

3

4

(1)

Total initial energy of two particles

$= \frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2}$

Total final energy of two particles

$= \frac{1}{2} m_{2} v_{2}^{2} + \frac{1}{2} m_{1} v_{1}^{2} + ε$

Using energy conservation principle,

$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2} + ε$

$∴$ $\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} - ε = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$

Topic Question Set

Q 1 :
Two bodies A and B of same mass undergo completely inelastic one dimensional collision. The body A moves with velocity $v_{1}$ while body B is at rest before collision. The velocity of the system after collision is $v_{2}$ . The ratio $v_{1} : v_{2}$ is [2024]

Q 2 :
A bullet of mass $m$ hits a block of mass $M$ elastically. The transfer of energy is the maximum, when: [2023]

Q 3 :
Body A of mass $4 m$ moving with speed $u$ collides with another body B of mass $2 m$ , at rest. The collision is head-on and elastic in nature. After the collision, the fraction of energy lost by the colliding body A is [2019]

Q 4 :
A moving block having mass $m$ , collides with another stationary block having mass $4 m$ . The lighter block comes to rest after collision. When the initial velocity of the lighter block is $v$ , then the value of coefficient of restitution (e) will be [2018]

Q 6 :
Two identical balls A and B having velocities of $0.5$ ${ms}^{- 1}$ and $- 0.3 {ms}^{- 1}$ respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will be [2016, 1994, 1991]

Q 7 :
Two particles A and B, move with constant velocities ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ . At the initial moment their position vectors are ${\vec{r}}_{1}$ and ${\vec{r}}_{2}$ respectively. The condition for particles A and B for their collision is [2015]

Q 8 :
A ball is thrown vertically downwards from a height of 20 m with an initial velocity $v_{0}$ . It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity $v_{0}$ is (Take $g = 10 {ms}^{- 2}$ ) [2015]

Q 10 :
Two particles of masses $m_{1}, m_{2}$ move with initial velocities $u_{1}$ and $u_{2}$ . On collision, one of the particles gets excited to a higher level, after absorbing energy $ε$ . If final velocities of particles be $v_{1}$ and $v_{2}$ , then we must have [2015]

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Topic Question Set

Q 1 : Two bodies A and B of same mass undergo completely inelastic one dimensional collision. The body A moves with velocity v1 while body B is at rest before collision. The velocity of the system after collision is v2. The ratio v1:v2 is [2024]

Q 2 : A bullet of mass m hits a block of mass M elastically. The transfer of energy is the maximum, when: [2023]

Q 3 : Body A of mass 4m moving with speed u collides with another body B of mass 2m, at rest. The collision is head-on and elastic in nature. After the collision, the fraction of energy lost by the colliding body A is [2019]

Q 4 : A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be [2018]

Q 6 : Two identical balls A and B having velocities of 0.5ms-1 and -0.3ms-1 respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will be [2016, 1994, 1991]

Q 7 : Two particles A and B, move with constant velocities v→1 and v→2. At the initial moment their position vectors are r→1 and r→2 respectively. The condition for particles A and B for their collision is [2015]

Q 8 : A ball is thrown vertically downwards from a height of 20 m with an initial velocity v0. It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity v0 is (Take g=10ms-2) [2015]

Q 10 : Two particles of masses m1,m2 move with initial velocities u1 and u2. On collision, one of the particles gets excited to a higher level, after absorbing energy ε. If final velocities of particles be v1 and v2, then we must have [2015]

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Q 1 :
Two bodies A and B of same mass undergo completely inelastic one dimensional collision. The body A moves with velocity $v_{1}$ while body B is at rest before collision. The velocity of the system after collision is $v_{2}$ . The ratio $v_{1} : v_{2}$ is [2024]

Q 2 :
A bullet of mass $m$ hits a block of mass $M$ elastically. The transfer of energy is the maximum, when: [2023]

Q 3 :
Body A of mass $4 m$ moving with speed $u$ collides with another body B of mass $2 m$ , at rest. The collision is head-on and elastic in nature. After the collision, the fraction of energy lost by the colliding body A is [2019]

Q 4 :
A moving block having mass $m$ , collides with another stationary block having mass $4 m$ . The lighter block comes to rest after collision. When the initial velocity of the lighter block is $v$ , then the value of coefficient of restitution (e) will be [2018]

Q 6 :
Two identical balls A and B having velocities of $0.5$ ${ms}^{- 1}$ and $- 0.3 {ms}^{- 1}$ respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will be [2016, 1994, 1991]

Q 7 :
Two particles A and B, move with constant velocities ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ . At the initial moment their position vectors are ${\vec{r}}_{1}$ and ${\vec{r}}_{2}$ respectively. The condition for particles A and B for their collision is [2015]

Q 8 :
A ball is thrown vertically downwards from a height of 20 m with an initial velocity $v_{0}$ . It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity $v_{0}$ is (Take $g = 10 {ms}^{- 2}$ ) [2015]

Q 10 :
Two particles of masses $m_{1}, m_{2}$ move with initial velocities $u_{1}$ and $u_{2}$ . On collision, one of the particles gets excited to a higher level, after absorbing energy $ε$ . If final velocities of particles be $v_{1}$ and $v_{2}$ , then we must have [2015]