PHY 2053, Spring 2009, Quiz 4 — Whiting
1. An object with mass 4.00 kg rests on a frictionless horizontal table and is connected to a
cable that passes over a pulley and is then fastened to a hanging object with mass 3.00 kg.
a) Find the acceleration of each object.
For the mass on the horizontal surface, we have T = mH × a. For the mass hanging vertically,
we have T − mV × g = −mV × a. By subtracting the second equation from the first, we can
eliminate T , and obtain mV × g = (mH + mV ) × a. Solving for a we find:
a=
mV × g
3.00 × 9.8
= 4.20 m/s2 .
=
mH + mV
4.00 + 3.00
b) Find the tension in the cable.
We can substitute this result into the first equation to solve for T . We get:
T = mH × a = 4.00 × 4.20 = 16.80 N.
We can check this in the second equation. On the left we get:
T − mV × g = 16.80 − 3.00 × 9.8 = 16.80 − 29.40 = −12.60 N,
while the right side also gives us: −mV × a = −3.00 × 4.20 = −12.60 N.
2. After falling from rest from a height of 19.60 m, a 1.00 kg ball rebounds upward reaching
a height of 4.90 m. The contact between the ball and the ground lasts 2.45 ms.
a) What is the average acceleration of the ball while in contact with the ground?
Use va2 = v02 +2a×∆x, with v0 = 0 to find the velocity with which it arrives at the ground:
√
√
va = − 2a × ∆x = − 2 ×−9.8 ×−19.60 = −19.60 m/s. Use 0 = vb2 + 2a × ∆x, with ∆x =
√
√
4.9 m, to find the velocity after the bounce: vb = −2a × ∆x = − −2 ×−9.8 × 4.90 =
9.80 m/s. Then the average acceleration during the bounce is given by:
a=
vf − vi
v − va
9.80 − (−19.60)
= b
=
= 12, 000 m/s2 .
∆t
t
0.00245
b) What average force (magnitude and direction) was exerted on the ball by the ground?
Newton’s Second Law gives us: F = m × a = 1.00 × 12, 000 = 12, 000 N, which is directed
vertically upward.