Question

a superball with mass m equal to 50 grams is dropped from a height of hi = 1.5 m . It collides with a table, then bounces up to a height of hf= 1.0m. The durationof the collision (the time during which the superball is in contact with the table) is tc = 15 m/s. In this problem, take the positive y direction to be upward,and use g = 9.8 m/s^2 for the magnitude of the acceleration due to gravity. Neglect air resistance.
1.Find the y component of the momentum, p before y, of the ball immediately before the collision.
2.Find the y component of the momentum of the ball immediately after the collision, that is, just as it is leaving the table.
3.Find the y component of the time-averaged force F avg,y, in newtons, that the table exerts on the ball.
4.Find Jy, the y component of the impulse imparted to the ball during the collision.
5.Find K after- K before, the change in the kinetic energy of the ball during the collision, in joules.

Answer 1

A ball of mass \(\mathrm{m}=50 \mathrm{~g}=0.05 \mathrm{~kg}\) is dropped from height \(h_{i}=1.5 \mathrm{~m}\).

The final height of the ball is \(h_{f}=1.0 \mathrm{~m}\).

The y component of the momentum, pbefore,y, of the ball immediately before the collision can be determined as follows:

According to law of conservation of energy initial potential energy is equal to final kinetic

energy.

$$ \begin{array}{l} m g h_{i}=(1 / 2) m v_{i}^{2} \\ g h_{i}=(1 / 2) v_{i}^{2} \Rightarrow v_{i}=\sqrt{2 g h_{i}}=\sqrt{2\left(9.8 m / s^{2}\right)(1.5 m)}=5.42 \mathrm{~m} / \mathrm{s} \end{array} $$

Therefore, the momentum of the ball is \(p=(0.05 \mathrm{~kg})(5.42 \mathrm{~m} / \mathrm{s})=0.271 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)

The velocity is in downward direction, thus, the momentum is \(p=-0.271 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)

(2) The y component of the momentum of the ball immediately after the collision, that is, just as it is leaving the table can be determined as follows:

The final height of the ball is \(h_{f}=1.0 \mathrm{~m}\).

According to law of conservation of energy initial potential energy is equal to final kinetic energy.

\(m g h_{f}=(1 / 2) m v_{f}^{2}\)

$$ g h_{f}=(1 / 2) v_{f}^{2} \Rightarrow v_{f}=\sqrt{2 g h_{f}}=\sqrt{2\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)(1 \mathrm{~m})}=4.43 \mathrm{~m} / \mathrm{s} $$

Therefore, the momentum of the ball is \(p=(0.05 \mathrm{~kg})(4.43 \mathrm{~m} / \mathrm{s})=0.22 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)

(3) Jy, the y component of the impulse imparted to the ball during the collision can be determined as follows:

The impulse is defined as change in momentum.

$$ J=p_{f}-p_{i}=0.22 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}-(-0.27 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s})=0.49 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s} $$

(4) The y component of the time-averaged force Favg,y, in newtons, that the table exerts on the ball can be determined as follows:

The impulse of the ball can also be defined as: product of average force and time.

$$ J=F_{\text {avg }} t $$

Therefore, the average force is,

$$ F_{\text {avg }}=J / t=\frac{0.49 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}}{15 \times 10^{-3} \mathrm{~s}}=32.67 \mathrm{~N} $$

(5) The change in the kinetic energy of the ball during the collision is,

$$ \begin{array}{l} \Delta K=(1 / 2) m v_{f}^{2}-(1 / 2) m v_{i}^{2} \\ \Delta K=(1 / 2) m\left(v_{f}^{2}-v_{i}^{2}\right)=(1 / 2)(0.05 k g)\left((4.43 \mathrm{~m} / \mathrm{s})^{2}-(5.42 \mathrm{~m} / \mathrm{s})^{2}\right) \end{array} $$

Therefore, the change in kinetic energy is,

$$ \Delta K=-0.243 J $$