Conservation of Momentum and Mechanical Energy

Q 1 :

A body of mass 1000 kg is moving horizontally with a velocity 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is [2024]

6
2
3
5

1

2

3

4

(4)

Momentum will remain conserve

$1000 \times 6 = 1200 \times v$

$v = 5 m / s$

Q 2 :

A stationary particle breaks into two parts of masses $m_{A}$ and $m_{B}$ , which move with velocities $v_{A}$ and $v_{B}$ respectively. The ratio of their kinetic energies $(K_{B} : K_{A})$ is [2024]

$m_{B} : m_{A}$
$1 : 1$
$m_{B} v_{B} : m_{A} v_{A}$
$v_{B} : v_{A}$

1

2

3

4

(4)

By conservation of linear momentum

$0 = m_{A} \vec{V_{A}} + m_{B} \vec{V_{B}}$

$\Rightarrow m_{A} \vec{V_{A}} = - m_{B} \vec{V_{B}}$

$\Rightarrow m_{A} V_{A} = m_{B} V_{B}$

$K = \frac{P^{2}}{2 m} \Rightarrow K \propto \frac{1}{m}$

$\Rightarrow \frac{K_{B}}{K_{A}} = \frac{m_{A}}{m_{B}} \Rightarrow \frac{K_{B}}{K_{A}} = \frac{v_{B}}{v_{A}}$

Q 3 :

A nucleus at rest disintegrates into two smaller nuclei with their masses in the ratio of 2:1. After disintegration, they will move [2024]

in opposite directions with speed in the ratio of 1:2 respectively
in opposite directions with speed in the ratio of 2:1 respectively
in the same direction with the same speed.
in opposite directions with the same speed.

1

2

3

4

(1)

From conservation of momentum

$P_{i} = P_{f}$

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

$\frac{v_{1}}{v_{2}} = \frac{2}{1},$ move in opposite direction with speed ratio 1 : 2.

Q 4 :

Consider two blocks A and B of masses $m_{1}$ = 10 kg and $m_{2}$ = 5 kg that are placed on a frictionless table. The block A moves with a constant speed v = 3 m/s towards the block B kept at rest. A spring with spring constant k = 3000 N/m is attached with the block B as shown in the figure. After the collision, suppose that the blocks A and B, along with the spring in constant compression state, move together, then the compression in the spring is, (Neglect the mass of the spring) [2025]

0.2 m
0.4 m
0.1 m
0.3 m

1

2

3

4

(3)

Conservation of Linear momentum,

$m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v_{cm}$

$v_{cm} \Rightarrow \frac{10 \times 3}{10 + 5} \Rightarrow \frac{30}{15} = 2 m / s$

Conservation of mechanical energy,

$\frac{1}{2} k x^{2} = \frac{1}{2} (10) {(3)}^{2} - [\frac{1}{2} (15) {(2)}^{2}]$

$\Rightarrow 3000 x^{2} = 90 - 60 = 30$

$\Rightarrow x^{2} = \frac{30}{3000} = \frac{1}{100}$

$\Rightarrow x = \frac{1}{10} m$

Q 5 :

A machine gun of mass 10 kg fires 20 g bullets at the rate of 180 bullets per minute with a speed of 100 ${ms}^{- 1}$ each. The recoil velocity of the gun is [2023]

0.02 m/s
2.5 m/s
1.5 m/s
0.6 m/s

1

2

3

4

(4)

$20 \times 10^{- 3} \times \frac{180}{60} \times 100 = 10 V$

$\Rightarrow v = 0.6 m/s$

Q 6 :

100 balls each of mass $m$ moving with speed $v$ simultaneously strike a wall normally and reflected back with same speed, in time $t$ s. The total force exerted by the balls on the wall is [2023]

$\frac{100 m v}{t}$
$\frac{200 m v}{t}$
$200 m v t$
$\frac{m v}{100 t}$

1

2

3

4

(2)

${\vec{P}}_{i} = N m v \hat{i}, {\vec{P}}_{f} = - N m v \hat{i}$

N is the number of balls striking the wall

[IMAGE 41]------------------

$N = 100$

$Δ \vec{P} = {\vec{P}}_{f} - {\vec{P}}_{i} = - 2 N m v \hat{i} = - 200 N m v \hat{i}$

${\vec{F}}_{total} = \frac{Δ \vec{P}}{Δ t} = - \frac{200 m v t}{t}$

$| \vec{F} |$ $= \frac{200 m v}{t}$

Q 7 :

The figure represents the momentum--time ( $p - t$ ) curve for a particle moving along an axis under the influence of a force. Identify the regions on the graph where the magnitude of the force is maximum and minimum respectively, if $(t_{3} - t_{2}) < t_{1}$ . [2023]

[IMAGE 42]

c and a
b and c
c and b
a and b

1

2

3

4

(3)

$| \frac{d \vec{p}}{d t} |$ $=$ $| \vec{F} |$ $\Rightarrow \frac{d \vec{p}}{d t} = slope of curve$

$Maximum slope: (c)$

$Minimum slope: (b)$

Q 8 :

A body of mass 5 kg is moving with a momentum of 10 kg ${ms}^{- 1}$ . Now a force of 2 N acts on the body in the direction of its motion for 5 s. The increase in the kinetic energy of the body is ________ J. [2023]

0%

(30)

$Given: M = 5 kg$

$P_{i} = 10 kg m/s (initial momentum)$

$Impulse = F Δ t = Δ P = P_{f} - P_{i}$

$\Rightarrow 2 \times 5 = P_{f} - 10$

$P_{f} = 20 kg m/s (final momentum)$

$Increase in KE = K E_{f} - K E_{i}$

$= \frac{P_{f}^{2}}{2 m} - \frac{P_{i}^{2}}{2 m} = \frac{400}{2 \times 5} - \frac{100}{2 \times 5} = 40 - 10 = 30 J$

Topic Question Set

Q 1 :

A body of mass 1000 kg is moving horizontally with a velocity 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is [2024]

Q 2 :

A stationary particle breaks into two parts of masses $m_{A}$ and $m_{B}$ , which move with velocities $v_{A}$ and $v_{B}$ respectively. The ratio of their kinetic energies $(K_{B} : K_{A})$ is [2024]

Q 3 :

A nucleus at rest disintegrates into two smaller nuclei with their masses in the ratio of 2:1. After disintegration, they will move [2024]

Q 5 :

A machine gun of mass 10 kg fires 20 g bullets at the rate of 180 bullets per minute with a speed of 100 ${ms}^{- 1}$ each. The recoil velocity of the gun is [2023]

Q 6 :

100 balls each of mass $m$ moving with speed $v$ simultaneously strike a wall normally and reflected back with same speed, in time $t$ s. The total force exerted by the balls on the wall is [2023]

Q 7 :

The figure represents the momentum--time ( $p - t$ ) curve for a particle moving along an axis under the influence of a force. Identify the regions on the graph where the magnitude of the force is maximum and minimum respectively, if $(t_{3} - t_{2}) < t_{1}$ . [2023]

[IMAGE 42]

Q 8 :

A body of mass 5 kg is moving with a momentum of 10 kg ${ms}^{- 1}$ . Now a force of 2 N acts on the body in the direction of its motion for 5 s. The increase in the kinetic energy of the body is ________ J. [2023]

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