Two vehicles are approaching an intersection. One is a 2900-kg pickup traveling
ID: 1355649 • Letter: T
Question
Two vehicles are approaching an intersection. One is a 2900-kg pickup traveling at 11.1 m/s from east to west (the x-direction), and the other is a 1325-kg sedan going from south to north (the +y-direction at 20.4 m/s).
(a) Find the x- and y-components of the net momentum of this system.
(b) What are the magnitude and direction of the net momentum?
PART TWO:
Force of a Baseball Swing. A baseball has mass 0.145 kg.
(a) If the velocity of a pitched ball has a magnitude of 36.0 m/s and the batted ball's velocity is 58.0 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.
13.68 kg·m/s (change in momentum)
13.68 kg·m/s (impulse)
(b) If the ball remains in contact with the bat for 2.00 ms, find the magnitude of the average force applied by the bat.
Explanation / Answer
1)
Let
m1 = 2900 kg
v1x = -11.1 m/s
m2 = 1325 kg
v2y = 20.4 m/s
Px = m1*v1x
= 2900*(-11.1)
= -32190 kg.m/s
Py = m2*v2y
= 1325*20.4
= 27030 kg.m/s
b) magnitude, P = sqrt(Px^2 + Py^2)
= sqrt(32190^2 + 27030^2)
= 42033.5 kg.m/s
direction : theta = tan^-1(Py/Px)
= tan^-1(27030/32190)
= 40 degrees west of north
2)
a) change in momentum = m*(v2 - v1)
= 0.145*(58 - (-36))
= 13.63 kg.m/s
we know, impulse = change in momentum
= 13.63 kg.m/s
b) Impulse = F*dt
==> F = impulse/dt
= 13.63/(2*10^-3)
= 6815 N
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