A ball with a mass of 275 g is dropped from rest, hits the floor, and rebounds u

Question

A ball with a mass of 275 g is dropped from rest, hits the floor, and rebounds upward. If the ball hits the floor with a speed of 2.60 m/s, rebounds with a speed of 1.84 m/s, and is in contact with the floor for 1.40 ms,determine the following.

A ball with a mass of 275 g is dropped from rest, hits the floor, and rebounds upward. If the ball hits the floor with a speed of 2.60 m/s, rebounds with a speed of 1.84 m/s, and is in contact with the floor for 1.40 ms,determine the following. (a) magnitude of the change in the ball's momentum (Let up be in the positive j hat direction.) _______________kg ? m/s (b) change in the magnitude of the ball's momentum (Let negative values indicate a decrease in magnitude.) _____________kg ? m/s (c) average force the floor exerts on the ball during their interaction (Use vector notation.) Favg = ________________ N

Explanation / Answer

Before the bounce, the ball is moving downward, so let's say its velocity is -2.6m/s (I used a negative number to indicate downward motion). So it's momentum is 0.275kg * (-2.6m/s) = -0.715 kg m/s (using a negative number to indicate downward momentum).

After the bounce, the ball's momentum is 0.275kg * 1.84 m/s = 0.506 kg m/s (now using positive numbers to indicate upward velocity/momentum)

A) Change in ball's momentum is final momentum minus initial momentum:
0.506 - (-0.715) = 1.221 kg m/s

B) Change in magnitude of the ball's momentum is:
|0.506| - |-0.715|
= 0.506 - 0.715
= -0.209kg m/s

C) The number in A makes more sense. When the ball bounced, first its downward momentum was reduced from 0.715 to zero, and then its upward momentum increased from zero to 0.506. Therefore, the total change was 1.221.

F = impulse / time

= 1.221 / (1.4 * 10^-3)

Force= 872.14 N

- - - Also to check answer for Part C- - -

The number in B is meaningless -- why find the difference (in magnitudes) between two (vector) quantities that have different (in this case, opposite) directions? Suppose that the ball's velocity changed from 1m/s down to 1m/s up. Do you subtract 1 - 1 to get zero, and therefore conclude that no acceleration took place? That's wrong, because you neglected to include the DIRECTION (up/down) of the vectors. You need to subtract 1UP from 1DOWN to get 2UP, and that's the real change in velocity. Similarly, in this case, you need to subtract 0.715DOWN from 0.506UP to get the real change in momentum. Just subtracting the numbers but ignoring the direction is pointless.

Force = mass * acceleration
Acceleration = change in velocity time
Force = (mass * change in velocity) time=  impulse / time
Change in momentum = (mass * change in velocity)

Change is momentum is directly related to the force acting on the ball.

= 0.275(2.6 + 1.84)1.4x10^-3

= 872.1 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.