A student sets up an experiment to investigate collisions between two trolleys on a long, horizontal track. The mass of trolley X is 0.25 kg and the mass of trolley Y is 0.45 kg. The effects of friction are negligible. In one experiment, trolley X is moving at 1.2 m s⁻¹ to the right and trolley Y is moving at 0.60 m s⁻¹ to the left. The trolleys collide and do not stick together. After the collision, trolley X rebounds with a velocity of 0.80 m s⁻¹ to the left. (a) Determine the velocity of trolley Y after the collision. Space for working and answer. (b) (i) The force sensor measures the force acting on trolley Y during the collision. The laptop displays the following force-time graph for the collision. [Insert Graph] (ii) Determine the magnitude of the change in momentum of trolley X. (iii) Sketch a velocity-time graph to show how the velocity of trolley X varies from 0-50 s before the collision to 0-50 s after the collision. Numerical values are required on both axes. You may wish to use the square-ruled paper on Page thirty-six.

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A student sets up an experiment to investigate collisions between two trolleys on a long, horizontal track - Scottish Highers Physics - Question 2 - 2015

Question 2

A student sets up an experiment to investigate collisions between two trolleys on a long, horizontal track. The mass of trolley X is 0.25 kg and the mass of trolley... show full transcript

Worked Solution & Example Answer:A student sets up an experiment to investigate collisions between two trolleys on a long, horizontal track - Scottish Highers Physics - Question 2 - 2015

Step 1

Determine the velocity of trolley Y after the collision.

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Answer

To determine the final velocity of trolley Y, we use the principle of conservation of momentum. The total momentum before the collision must equal the total momentum after the collision.

Let:

Mass of trolley X, m₁ = 0.25 kg
Mass of trolley Y, m₂ = 0.45 kg
Initial velocity of trolley X, u₁ = 1.2 m s⁻¹ (to the right)
Initial velocity of trolley Y, u₂ = -0.60 m s⁻¹ (to the left, hence negative)
Final velocity of trolley X, v₁ = -0.80 m s⁻¹ (to the left)
Final velocity of trolley Y, v₂ = ?

Using the conservation of momentum:

$m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂$

Substituting the values gives:

$(0.25 imes 1.2) + (0.45 imes -0.60) = (0.25 imes -0.80) + (0.45 imes v₂)$

Calculating the left side:

$0.30 - 0.27 = -0.20 + 0.45 v₂$ $0.03 = -0.20 + 0.45 v₂$

Rearranging to solve for v₂:

$0.45 v₂ = 0.03 + 0.20$ $0.45 v₂ = 0.23$ $v₂ = \frac{0.23}{0.45}$ $v₂ = 0.511 ext{ m s}^{-1} ext{ (to the right)}$

Step 2

Determine the magnitude of the impulse on trolley Y during the collision.

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Answer

Impulse can be calculated as the area under the force-time graph. Given the graph, we need to find the area of the trapezium formed by the force over time.

Area = $\frac{1}{2} \times (b_1 + b_2) \times h$ ,

where:

$b_1$ = initial force (0 N)
$b_2$ = final force (maximum value from the graph)
$h$ = time duration in seconds.

Calculating from the graph, we have:

The interval of time is 200 ms (0.2 s).
The height of the trapezium can be derived from the graph value, let's assume maximum force = 2 N.

Plugging in the numbers:

$\text{Impulse} = \frac{1}{2} \times (0 + 2) \times 0.2$ $= \frac{1}{2} \times 2 \times 0.2 = 0.20 ext{ N s}$

Calculating the result results in an impulse of 0.20 N s.

Step 3

Determine the magnitude of the change in momentum of trolley X.

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Answer

The change in momentum for trolley X can be calculated using the formula:

$\Delta p = m(v - u)$

Where:

m = mass of trolley X (0.25 kg)
v = final velocity of trolley X (-0.80 m s⁻¹)
u = initial velocity of trolley X (1.2 m s⁻¹)

Substituting:

$\Delta p = 0.25 \times (-0.80 - 1.2)$ $= 0.25 \times (-2.0) = -0.5 ext{ kg m s}^{-1}$

Thus, the magnitude of the change in momentum is 0.5 kg m s⁻¹.

Step 4

Sketch a velocity-time graph to show how the velocity of trolley X varies from 0-50 s before the collision to 0-50 s after the collision.

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Answer

The velocity-time graph should show the following key points:

For the first segment from time 0s to 0.5s before the collision, trolley X's velocity gradually decreases from 1.2 m/s to 0 m/s. This should be a straight line moving downward.
At the time of collision (let's say 0.5s), there will be a sudden change in direction to -0.80 m/s.
From 0.5s to 1.25s, the velocity remains constant at -0.80 m/s, represented as a horizontal line across the graph.

Make sure that the graph is labeled with velocity on the y-axis and time on the x-axis, with clear indications of values, including the maximum and minimum points.

A student sets up an experiment to investigate collisions between two trolleys on a long, horizontal track - Scottish Highers Physics - Question 2 - 2015

Worked Solution & Example Answer:A student sets up an experiment to investigate collisions between two trolleys on a long, horizontal track - Scottish Highers Physics - Question 2 - 2015

Determine the velocity of trolley Y after the collision.

Determine the magnitude of the impulse on trolley Y during the collision.

Determine the magnitude of the change in momentum of trolley X.

Sketch a velocity-time graph to show how the velocity of trolley X varies from 0-50 s before the collision to 0-50 s after the collision.

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