Solved: Two players are engaged in a game of table tennis. Player A strikes the ball at... [IB Physics (New syllabus)]

Question

SLPaper 2

Two players are engaged in a game of table tennis. Player A strikes the ball at a height of 0.27 m above the table's surface, measured from the tabletop to the bottom of the ball. The ball's initial velocity is 13.5 m/s in the horizontal direction.

Assume that air resistance is negligible.

The ball bounces and then reaches a peak height of 0.19 m above the table with a horizontal speed of 11.2 m s $^{-1}$ . The mass of the ball is 3.6 g.

Show that the time taken for the ball to reach the surface of the table is about 0.2 s.

[1]

Verified

Solution

$t = \sqrt{\frac{2d}{g}}$
$t = \sqrt{\frac{2 \times 0.27}{9.8}} = 0.23 s$ 1 mark

Answer to 2 or more significant figures or formula with variables replaced by correct values.

1 mark total

Sketch, on the axes rpovided below, a graph showing the variation with time of the vertical component of velocity $v_v$ of the ball until it reaches the table surface. Take $g$ to be $+10 \, \text{m s}^{-2}$

[2]

Verified

Solution

Increasing straight line from zero up to 0.2 s on the x-axis 1 mark
Line has a gradient of 10 1 mark

2 marks total

The net is stretched across the middle of the table. The table has a length of 4.84 m and the net has a height of 9.8 cm.

Show that the ball will go over the net.

[3]

Verified

Solution

$t = \frac{2.42}{13.5} = 0.179 \text{ s}$ 1 mark
$y = \frac{1}{2} \times 10 \times 0.179^2 = 0.160 \text{ m}$ 1 mark
$(0.27 - 0.160) = 0.11 > 0.098$ so it goes over the net 1 mark

Player B intercepts the ball at its highest point. Holding the paddle stationary in a vertical position, Player B makes contact with the ball for 0.010 s. Assume the collision is perfectly elastic.

Calculate the average force exerted by the ball on the paddle. State your answer to an appropriate number of significant figures.

[2]

Verified

Solution

$\Delta v = 22.4\, \text{m}\, \text{s}^{-1}$ 1 mark
$F = \frac{0.0036 \times 22.4}{0.01}$
$F = 8.06\, \text{N}$ 1 mark

2 marks total

Remember to state your answer to an appropriate number of significant figures as requested in the question

Determine the kinetic energy of the ball immediately after the bounce.

[3]

Verified

Solution

Use of $KE = \frac{1}{2}mv^2 + mgh$ 1 mark
Correct substitution: $KE = \frac{1}{2} \times 0.0036 \times 11.2^2 + 0.0036 \times 9.8 \times 0.19$ 1 mark
Final answer: $0.23$ J 1 mark

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