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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 hi= 1.0m. The duration of 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 average, y, in newtons, that the table exerts on the ball.

4.Find J, y, 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.

Relevant knowledge

Linear momentum can be described as the object’s mass multiplied by its velocity. It is often represented by the pp letter and is measured in kilogramm/s (kgm/s).

*p=mv*

In a closed system (where there are no external forces that affect any of the objects), the total momentum is conserved. If we add up every object’s momentum at one moment in time, then add them up again at a later moment, the two sums must be the same (even if individual objects’ momenta change). In other words, the initial momentum must be equal to final momentum.

*p _{i}=p_{f}*

However, if external forces act on the system, the total momentum will change and we cannot use this equation.

A ball of mass [katex]\mathrm{m}=50 \mathrm{~g}=0.05 \mathrm{~kg}[/katex]

The ball’s final height is [katex]h_{f}=1.0 \mathrm{~m}[/katex]

You can determine the y component of momentum, pbefore.y, by looking at the ball just before collision:

The law of conservation energy states that initial potential energy equals final kinetic energy.

energy.

[katex] \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} [/katex]

The momentum of the ball therefore is [katex]p=(0.05 \mathrm{~kg})(5.42 \mathrm{~m} / \mathrm{s})=0.271 \mathrm{~kg}-\mathrm{m} / \mathrm{s}[/katex]

The velocity is decreasing, so momentum is [katex]p=-0.271 \mathrm{~kg}-\mathrm{m} / \mathrm{s}[/katex]

(2) The y-component of the momentum of a ball just after a collision, that is, as the ball is leaving the table, can be determined by the following:

The ball’s final height is [katex]h_{f}=1.0 \mathrm{~m}[/katex]

The law of conservation energy states that initial potential energy and final kinetic energy are equal.

[katex]m g h_{f}=(1 / 2) m v_{f}^{2}[/katex]

[katex] 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} [/katex]

The momentum of the ball therefore is [katex]p=(0.05 \mathrm{~kg})(4.43 \mathrm{~m} / \mathrm{s})=0.22 \mathrm{~kg}-\mathrm{m} / \mathrm{s}[/katex]

(3) Jy is the y component to the impact that was given to the ball by the collision. It can be calculated as follows:

An impulse can be defined as a change in momentum.

[katex] 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} [/katex]

(4) The time-averaged force Favg.y in newtons that the table exerts upon the ball can be calculated as follows:

The impulse of a ball can also be described as the product of average force, time and force.

[katex]h_{f}=1.0 \mathrm{~m}[/katex]0

The average force therefore is

[katex]h_{f}=1.0 \mathrm{~m}[/katex]1

(5) The ball’s kinetic energy changes during collision.

[katex]h_{f}=1.0 \mathrm{~m}[/katex]2

The change in kinetic energy therefore is.

[katex]h_{f}=1.0 \mathrm{~m}[/katex]3