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Collision

Practice Questions:

Question 1: moderate

A stationary body of mass m explodes into 3 parts with mass ratio of \(1 : 3 : 3\). The two fragments with equal mass move at right angles to each other with velocity of \(15\text{ ms}^{-1}\). The velocity of the third fragment is (in \(\text{ms}^{-1}\)):

1. \(15\sqrt{2}\)

2. 5

3. \(20\sqrt{2}\)

4. \(45\sqrt{2}\)

View Answer

The ratio of masses is \(m' : 3m' : 3m'\). The combined momentum of the two perpendicular \(3m'\) masses is \(P = \sqrt{(3m' \times 15)^2 + (3m' \times 15)^2} = 45\sqrt{2} m'\). Conservation of momentum requires the third fragment \(m'\) to balance this: \(m' v_3 = 45\sqrt{2} m' ⇒ v_3 = 45\sqrt{2}\text{ ms}^{-1}\).

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Question 2: moderate

A body with kinetic energy K moving in +X direction splits up into two parts A and B of equal mass on its own. Part ‘A’ moves back in -X direction with a velocity equal in magnitude to the initial velocity of the body. The kinetic energy of part B will be:

1. K

2. 4K

3. \(\frac{K}{2}\)

4. \(\frac{9}{2}K\)

View Answer

Let the total mass be \(2m\), so \(K = mv_0^2\). Under momentum conservation, \(2mv_0 = m(-v_0) + mv_B ⇒ v_B = 3v_0\). The kinetic energy of part B is \(K_B = \frac{1}{2}m(3v_0)^2 = \frac{9}{2}mv_0^2 = \frac{9}{2}K\).

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Question 3: moderate

A ball of mass \( m \) approaches a wall of mass \( M \) (\( M \gg m \)) with speed \( 4\text{ m/s} \) along the normal to the wall. The speed of the wall is \( 1\text{ m/s} \) towards the ball. The speed of the ball after an elastic collision with the wall is:

1. \( 5\text{ m/s} \) away from the wall

2. \( 9\text{ m/s} \) away from the wall

3. \( 3\text{ m/s} \) away from the wall

4. \( 6\text{ m/s} \) away from the wall

View Answer

Using the coefficient of restitution \( e = 1 \), the relative velocity of separation equals the relative velocity of approach. The approach velocity is \( 4 - (-1) = 5\text{ m/s} \). After collision, the relative separation velocity is \( v' - 1 = 5 \implies v' = 6\text{ m/s} \) away from the wall.

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Question 4: moderate

A block of mass \( m \) moving with a velocity \( v \) collides with another block of mass \( M \) at rest. The two blocks stick together due to the collision. The loss of K.E. expressed as a fraction of total initial kinetic energy is:

1. \( \frac{M}{m+M} \)

2. \( \frac{m}{m+M} \)

3. \( \frac{M^2}{m+M} \)

4. \( \frac{M-m}{m+M} \)

View Answer

By conservation of momentum, the final velocity after a completely inelastic collision is \( v_f = \frac{mv}{m+M} \). The fractional loss of kinetic energy is \( \frac{K_i - K_f}{K_i} = 1 - \frac{\frac{1}{2}(m+M)v_f^2}{\frac{1}{2}mv^2} = \frac{M}{m+M} \).

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