A white snooker ball and a black snooker ball travel towards each other in a straight line. The white ball and the black ball each have a mass of 0.180 kg. Just before the balls collide head-on, the white ball is travelling at 2.60 m s⁻¹ to the right and the black ball is travelling at 1.80 m s⁻¹ to the left. After the collision, the black ball rebounds with a velocity of 2.38 m s⁻¹ to the right. (a) (i) Determine the velocity of the white ball immediately after the collision. Space for working and answer (ii) The collision between the balls is inelastic. State what is meant by an inelastic collision. (b) A student carries out an experiment to measure the average force exerted by a cue on a ball. The cue hits the stationary ball. The timer records the time the cue is in contact with the ball. The computer displays the speed of the ball. The results are shown. Time of contact between the cue and the ball = (0.040 - 0.001) s Speed of the ball immediately after contact = (0.84 ± 0.01) m s⁻¹ Mass of the ball = (0.180 ± 0.001) kg. (i) Calculate the average force exerted on the ball by the cue. An uncertainty in this value is not required. Space for working and answer (ii) Determine the percentage uncertainty in the value for the average force on the ball.

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A white snooker ball and a black snooker ball travel towards each other in a straight line - Scottish Highers Physics - Question 2 - 2017

Question 2

A white snooker ball and a black snooker ball travel towards each other in a straight line. The white ball and the black ball each have a mass of 0.180 kg. Just be... show full transcript

Worked Solution & Example Answer:A white snooker ball and a black snooker ball travel towards each other in a straight line - Scottish Highers Physics - Question 2 - 2017

Step 1

(i) Determine the velocity of the white ball immediately after the collision.

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Answer

To find the white ball's velocity after the collision, we start with the principle of conservation of momentum:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

Where:

$m_1 = 0.180 \, kg$ (mass of white ball)
$u_1 = 2.60 \, m/s$ (initial velocity of white ball)
$m_2 = 0.180 \, kg$ (mass of black ball)
$u_2 = -1.80 \, m/s$ (initial velocity of black ball, negative because it's to the left)
$v_2 = 2.38 \, m/s$ (final velocity of black ball)

Substituting the values, we get:

$0.180 imes 2.60 + 0.180 imes (-1.80) = 0.180 v_1 + 0.180 imes 2.38$

Simplifying:

$0.468 - 0.324 = 0.180 v_1 + 0.428$

This results in:

$0.144 = 0.180 v_1 + 0.428$

Now, isolating $v_1$ gives:

$0.180 v_1 = 0.144 - 0.428$ $0.180 v_1 = -0.284$

Thus, $v_1 = -1.58 \ m/s$

The velocity of the white ball immediately after the collision is approximately $1.58 \, m/s \text{ to the left}$ .

Step 2

(ii) The collision between the balls is inelastic. State what is meant by an inelastic collision.

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Answer

An inelastic collision is defined as one in which the total kinetic energy of the system is not conserved during the collision. This means that some of the kinetic energy is transformed into other forms of energy, such as heat or sound, or it could be used to deform the objects involved. In an inelastic collision, the objects may stick together or not separate completely after the collision.

Step 3

(i) Calculate the average force exerted on the ball by the cue.

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Answer

The average force exerted by the cue can be calculated using the equation:

$F = \frac{\Delta p}{\Delta t}$

Where:

$\Delta p$ is the change in momentum
$\Delta t$ is the time of contact

The change in momentum is given by:

$\Delta p = m \Delta v$

Calculating $\Delta v$ :

$\Delta v = v_f - v_i = 0.84 \ m/s - 0 = 0.84 \ m/s$

Thus, $\Delta p = (0.180 \ kg) \times (0.84 \ m/s) = 0.1512 \ kg \, m/s$

The time of contact is:

$\Delta t = 0.040 \ s$

Now substituting these into the force equation: $F = \frac{0.1512 \ kg \, m/s}{0.040 \ s} = 3.78 \ N$

The average force exerted on the ball by the cue is approximately $3.78 \, N$ .

Step 4

(ii) Determine the percentage uncertainty in the value for the average force on the ball.

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Answer

To determine the percentage uncertainty in the average force calculation, we use the formula for percentage uncertainty:

$\text{Percentage Uncertainty} = \left( \frac{\Delta F}{F} \right) \times 100\%$

Given the average force $F = 3.78 \ N$ with an uncertainty of $\Delta F = 0.01 \ N$ , the percentage uncertainty is calculated as follows:

$\text{Percentage Uncertainty} = \left( \frac{0.01}{3.78} \right) \times 100\% \approx 0.26\%$

Thus, the percentage uncertainty in the value for the average force on the ball is approximately 0.26%.

A white snooker ball and a black snooker ball travel towards each other in a straight line - Scottish Highers Physics - Question 2 - 2017

Worked Solution & Example Answer:A white snooker ball and a black snooker ball travel towards each other in a straight line - Scottish Highers Physics - Question 2 - 2017

(i) Determine the velocity of the white ball immediately after the collision.

(ii) The collision between the balls is inelastic. State what is meant by an inelastic collision.

(i) Calculate the average force exerted on the ball by the cue.

(ii) Determine the percentage uncertainty in the value for the average force on the ball.

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