Iin physics laboratory a trolley of unknown mass ,m, moves along a track at a constant velocity of 0,4m-s¹. A wooden block,mass500g is released directly above of and on top of the trolley. The trolley, along with the block, continues moving at 0,15m-s¹ in the same direction.

Question

Iin physics laboratory a trolley of unknown mass ,m, moves along a track at a constant velocity of 0,4m-s¹. A wooden block,mass500g is released directly above of and on top of the trolley. The trolley, along with the block, continues moving at 0,15m-s¹ in the same direction.

1. Is the collision between the wooden block

and the trolley Elastic or Inelastic?Use a suitable calculation to justify the answer.
2. Draw a velocity versue time sketch graph to illustrate the motion of the trolley before and after
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Answer 1

1. To determine if the collision is elastic or inelastic, we can use the principle of conservation of momentum. In an elastic collision, both momentum and kinetic energy are conserved. In an inelastic collision, only momentum is conserved and kinetic energy may be lost.

Before the collision, the trolley is moving at a velocity of 0.4 m/s. After the collision, the trolley and the block move together at a velocity of 0.15 m/s.

To find the momentum before and after the collision, we can use the equation p = mv, where p is momentum, m is mass, and v is velocity.

Before the collision:
Momentum of the trolley = m_trolley * v_trolley = m_t * 0.4
Momentum of the block = m_block * 0 (since it is released from rest)

After the collision:
Momentum of the combined system (trolley + block) = (m_trolley + m_block) * 0.15

If the total momentum before the collision is equal to the total momentum after the collision, then the collision is elastic. If they are not equal, then it is an inelastic collision.

(m_t * 0.4) = (m_t + m_block) * 0.15

Simplifying the equation:

0.4m_t = 0.15m_t + 0.15m_block

0.25m_t = 0.15m_block

m_t = (0.15/0.25) * m_block

m_t = 0.6 * m_block

This means that the mass of the trolley is 0.6 times the mass of the block.

Since the momentum before the collision is not equal to the momentum after the collision, the collision is inelastic.

2. The velocity-time graph would look like this:

Velocity (m/s)
| A
| |
| |
|__________________|_______
0 t1 t2
Time

Before the collision (from 0 to t1): The trolley is moving at a constant velocity of 0.4 m/s.

At the start of the collision (at t1): The velocity abruptly changes to 0.15 m/s as the block is released on top of the trolley.

After the collision (from t1 to t2): The trolley and block move together at a constant velocity of 0.15 m/s.

Note: The graph should show a sudden change in velocity at t1 to represent the collision.