Momentum in a collision and the force it needs

Forces

GCSE Physics (8463)

ORIGINAL

Conservation of momentum for the velocity after, then force as the rate of change of momentum.

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Question

A trolley of mass

0.80\,\text{kg}

moving at

2.5\,\text{m/s}

collides with a stationary trolley of mass

1.2\,\text{kg}

. The trolleys stick together. (a) Calculate their velocity just after the collision. (b) The collision lasts

0.040\,\text{s}

. Calculate the average force on the

1.2\,\text{kg}

trolley.

Every step worked, with the reasoning.

1
Momentum before $= mv = 0.80 \times 2.5 + 1.2 \times 0 = 2.0\,\text{kg m/s}$
Momentum $p = mv$ ; the stationary trolley has zero momentum.
2
Momentum after $= (0.80 + 1.2)v = 2.0v$
The trolleys move off together, so use their combined mass.
3
$2.0v = 2.0 \Rightarrow v = 1.0\,\text{m/s}$
In a closed system, the total momentum before a collision equals the total momentum after it.
4
Change in momentum of the $1.2\,\text{kg}$ trolley $= m\Delta v = 1.2 \times (1.0 - 0) = 1.2\,\text{kg m/s}$
Use that trolley's own mass and its own change in velocity.
5
$F = \dfrac{m\Delta v}{\Delta t} = \dfrac{1.2}{0.040} = 30\,\text{N}$
Force equals the rate of change of momentum. The $0.80\,\text{kg}$ trolley loses the same $1.2\,\text{kg m/s}$ , so it feels $30\,\text{N}$ in the opposite direction.

Answer: (a) $1.0\,\text{m/s}$ in the original direction of motion; (b) $30\,\text{N}$