Page 104 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 8] MOMENTUM 89

When a force F acts on a body of mass m and changes its velocity from v 1 to v 2 , according to the second
law of motion

(v 2 − v 1 )
F = ma = m
t

Hence we have the impulse-momentum theorem:

Ft = p − p 1 = m(v 2 − v 1 )
2
Impulse = momentum change

The momentum change of the body is equal to the impulse provided by the force acting on it.

SOLVED PROBLEM 8.1
Find the momentum of a 50-kg boy running at 6 m/s.

p = mv = (50 kg)(6 m/s) = 300 kg·m/s

SOLVED PROBLEM 8.2
A 160-lb woman runs 1 mi in 5 min. What is her average momentum?

The woman’s mass m and average velocity ¯v are, respectively,

w 160 lb (1mi)(5280 ft/mi)
m = = = 5 slugs ¯ v = = 17.6 ft/s
g 32 ft/s 2 (5 min)(60 s/min)
Hence her average momentum is p = m ¯v = (5 slugs)(17.6 ft/s) = 88 slug·ft/s.

SOLVED PROBLEM 8.3
A 46-g golf ball is struck by a club and flies off at 70 m/s. If the head of the club was in contact with the
ball for 0.5 ms, what was the average force on the ball during the impact?

The ball started from rest, so v 1 = 0 and its momentum change is

m(v 2 − v 1 ) = mv 2 = (0.046 kg)(70 m/s) = 3.22 kg·m/s

Since 1 ms = 1 millisecond = 10 −3 s, here t = 0.5ms = 5 × 10 −4 s and

m(v 2 − v 1 ) 3.22 kg·m/s 3
F = = = 6.4 × 10 N = 6.4kN
t 5 × 10 −4 s
This is equal to 1450 lb. According to the third law of motion, a recoil force of the same magnitude but opposite
direction acts on the club’s head during the impact. No golf club could withstand so large a stationary load, but the
impact is so brief that it merely bends the shaft temporarily by a few centimeters.

SOLVED PROBLEM 8.4
A certain DC-9 airplane has a mass of 50,000 kg and a cruising velocity of 700 km/h. Its engines develop
a total thrust of 70,000 N. If air resistance, change in altitude, and fuel consumption are ignored, how
long does it take the airplane to reach its cruising velocity, starting from rest?