This preview shows pages 1–2. Sign up to view the full content.
Lesson 12 Homework Solutions
1.
After jumping vertically upward to make a slam dunk, a 70kg
basketball player is essentially in free fall as he returns to the floor. If
he falls through a distance of 1.5 m, what is his rate of change of
momentum as he returns to the floor?
The force on a 70kg player in free fall is his weight.
Weight = 70 kg
×
9.8 ms
2
= 686 N
Since rate of change of momentum is a measure of the force that
causes the change in momentum, the rate of change of momentum =
686 kgms
1
.
2.
If you are riding in a car traveling at 72 kmh
1
, what is your
momentum relative to (i) the ground? (ii) the car?
(i)
Your speed relative to the ground = 72 kmh
1
= 20 ms
1
If m is your mass, then the momentum relative to the ground is:
P = mv =
20v kgms
1
(ii)
Your speed relative to the car is zero. Therefore, momentum
relative to the car is
zero.
3.
A 6.0g bullet traveling at 200 ms
1
strikes a bullet proof vest and
comes to a stop in 500
µ
s. What is the average force exerted by the
bullet on the vest?
Initial velocity of the bullet is:
v
i
= 200 ms
1
Initial momentum P
i
= mv
i
= 0.006 kg
×
200 ms
1
= 1.2 kg ms
1
Final momentum P
f
= 0
The change in momentum
∆
P = P
i
– P
f
= 1.2 kg ms
1
6
P1
.
2
The rate of change of momentum =
2400 N
t
500 10
−
∆
==
∆×
The average force exerted by the bullet is a measure of its rate of
change of momentum.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 05/01/2011 for the course PHY 2048 taught by Professor George during the Fall '10 term at Edison State College.
 Fall '10
 George
 Physics, Momentum, Work

Click to edit the document details