A bullet of mass m B = 1.09 ? 10 ? 2 kg is moving with a speed of 101 . m / s wh
ID: 2264658 • Letter: A
Question
A bullet of mass mB=1.09?10?2kg is moving with a speed of 101.m/s when it collides with a rod of mass mR=6.51kg and length L=1.04m (shown in the figure). The rod is initially at rest, in a vertical position, and pivots about an axis going through its center of mass. The bullet embeds itself in the rod at a distance L4 from the pivot point. As a result, the bullet-rod system starts rotating.
a) Find the magnitude of the angular velocity, ? , of the bullet-rod system after the collision. You can neglect the width of the rod and can treat the bullet as a point mass.
b) What is the change in kinetic energy in the collision?
Explanation / Answer
A bullet of mass
mB = 0.01 kg
is moving with a speed of 129 m/s when it collides with a rod of mass
mR = 6 kg
and length
L = 2 m
(shown in the figure). The rod is initially at rest, in a vertical position, and pivots about an axis going through its center of mass. The bullet imbeds itself in the rod at a distance L/4 from the pivot point. As a result, the bullet-rod system starts rotating.
(a) Find the angular velocity, ?, of the bullet--rod system after the collision. You can neglect the width of the rod and can treat the bullet as a point mass.
(b) How much kinetic energy is lost in the collision?
For a point mass, angular momentum (L) = m * v * r. For anything more complex, you need the moment of inertia and L = I * rotational velocity.
Angular momentum must be conserved before and after the collision. Before, all the momentum is in the bullet.
L = m * v * r
L = .01kg * 129m/s * .5m (bullet hits L/4 or .5 m from axis)
L = .645 kg m^2/s
After the collision same must be true, but of the whole system.
L = (I(rod) + I(bullet)) * w
I for a rod about the center is 1/12 m L^2
I for point mass is m r^2
L = (1/12 * 6kg * (2m)^2 + .001kg * (.5m)^2) * w
.322 rad/s = w
KE(before) = 1/2 * m * v^2
KE(before) = 1/2 * .01kg * (129m/s)^2
KE(before) = 83.205J
KE (after) = 1/2 * (I(rod) + I(bullet)) * w^2
KE (after) = 1/2 * (1/12 * 6kg * (2m)^2 + .001kg * (.5m)^2) * (.322 rad/s)^2
KE (after) = .104 J
KE (diff) = 83.205J - .104J
KE (diff) = 83.101J
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.