A glob of putty thrown to the right bounces off a 0.700-kg cart initially at res
ID: 1433316 • Letter: A
Question
A glob of putty thrown to the right bounces off a 0.700-kg cart initially at rest on a low-friction track. The collision takes 0.15 s, and the coefficient of restitution is 0.76. The cart has a final velocity of 1.0 m/s to the right, and the glob has a final velocity of 0.834 m/s to the left. Suppose that the positive x direction is to the right. What is the inertia of the glob? What is the x component of the average acceleration of the cart during the collision? What is the x component of the average acceleration of the glob during the collision?
Explanation / Answer
Let m be the inertia of glob
e = 1+0.834/V
e = 0.76
V = initial velocity of glob
V = 2.413 m/sec
conserving momentum
m (2.413) = 0.7x1 -m (0.834)
m = 0.21557 kg
acceleration of cart
a = 1/0.15 = 6.67 m/sec^2
acceleration of glob
a =- 0.834-2.413 /0.15
a = -21.646m/sec^2
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.