The figure shows a 0.3 kg baseball just before and just after it collides with a
ID: 1504852 • Letter: T
Question
The figure shows a 0.3 kg baseball just before and just after it collides with a bat. Just before, the ball has velocity ModifyingAbove v With right-arrow Subscript 1 of magnitude 12.2 m/s and angle 1 = 39.3°. Just after, it is traveling directly upward with velocity ModifyingAbove v With right-arrow Subscript 2 of magnitude 10.7 m/s. The duration of the collision is 1.20 ms. What are the (a) magnitude and (b) direction (relative to the positive direction of the x axis) of the impulse on the ball from the bat? What are the (c) magnitude and (d) direction of the average force on the ball from the bat?
Explanation / Answer
given data
m=0.3 kg
v1 = 12.2 m/s
1 = 39.3°
v2 = 10.7 m/s
t=1.20ms = 1.20*10^3 s
What are the (a) magnitude and (b) direction (relative to the positive direction of the x axis) of the impulse on the ball from the bat?
Impulse equals change in momentum.
The baseball is initially moving with vi = (12.2 cos i 12 sin j) m/s.
After the collision, its vf = +10 j m/s.
We will calculate p in terms of x- and y-components, then use them to answer the questions
change in momentum Px = m*vx=0.3*(0-(-9.44)) = 2.832 N.s
change in momentum Py = m*vy=0.3*(10-(-7.72)) = 5.136N.s
a) P = sqrt(2.832^2+5.136^2) =5.86 N.s
b) angle = tan^-1 (5.136/2.832) = 61.12 deg
c) F(avg) = P/t = 5.86/0.0012 =4883 N along P
d) same direcion as P .