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A trolley of mass 2.0kg moving to the right at 0.3 ms-1 collided with a trolley of mass 4.0kg which is at rest. The 4.0kg trolley moves off to the right at 0.7ms-1. Show that the 2.0kg trolley rebounds and moves off at 1.1ms-1

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User FrankS
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Answer: To show that the 2.0 kg trolley rebounds and moves off at 1.1 m/s after the collision, we can use the principle of conservation of momentum, which states that the total momentum of a system of objects remains constant if no external forces act on it.

Let's denote the initial velocity of the 2.0 kg trolley as u1 = 0.3 m/s (to the right), the initial velocity of the 4.0 kg trolley as u2 = 0 m/s (at rest), the final velocity of the 2.0 kg trolley as v1 (to be determined), and the final velocity of the 4.0 kg trolley as v2 = 0.7 m/s (to the right).

The momentum of an object is given by the product of its mass and velocity: momentum = mass * velocity.

According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.

Total initial momentum = Total final momentum

(mass1 * initial velocity1) + (mass2 * initial velocity2) = (mass1 * final velocity1) + (mass2 * final velocity2)

Substituting the given values:

(2.0 kg * 0.3 m/s) + (4.0 kg * 0 m/s) = (2.0 kg * v1) + (4.0 kg * 0.7 m/s)

0.6 kg m/s = 2.0 kg * v1 + 2.8 kg m/s

Rearranging the equation to solve for v1:

2.0 kg * v1 = 0.6 kg m/s - 2.8 kg m/s

2.0 kg * v1 = -2.2 kg m/s

v1 = -2.2 kg m/s / 2.0 kg

v1 ≈ -1.1 m/s

So, the final velocity of the 2.0 kg trolley after the collision is approximately -1.1 m/s (to the left), which indicates that the 2.0 kg trolley rebounds and moves off to the left at 1.1 m/s.

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User Gallymon
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