Help 1.A railroad car of mass 2.56 10 4 kg is moving with a speed of 4.16 m/s. I
ID: 1541714 • Letter: H
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1.A railroad car of mass 2.56 104 kg is moving with a speed of 4.16 m/s. It collides and couples with three other coupled railroad cars, each of the same mass as the single car and moving in the same direction with an initial speed of 2.08 m/s.
What is the speed of the four cars after the collision? (Round your answer to at least two decimal places.) ……M/S
(b) How much mechanical energy is lost in the collision? …… J
2. A tennis ball of mass 57.0 g is held just above a basketball of mass 637 g. With their centers vertically aligned, both balls are released from rest at the same time, to fall through a distance of 1.15 m, as shown in the figure below.
(a) Find the magnitude of the downward velocity with which the basketball reaches the ground.
m/s
(b) Assume that an elastic collision with the ground instantaneously reverses the velocity of the basketball while the tennis ball is still moving down. Next, the two balls meet in an elastic collision. To what height does the tennis ball rebound?
m
3. As shown below, a bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length L and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle? (Use any variable or symbol stated above along with the following as necessary: g.)
V=?
Explanation / Answer
Apply conservation of momentum
If M is the mass of a single car, conservation of momentum gives
( M+ 3M) vf = M ( 4.16) ) + 3M ( 2.08)
4M vf = M ( 4.16 + 3 ( 2.08))
vf= 2.6 m/s
(b)
The kinetic energy lost is
KE_ totl = KE_i - KE_f
= 1/2 * M ( 4.16)^2 + 1/2 ( 3M) ( 2.08)^2 - 1/2 * ( 4M) ( 2.6)^2
= M ( 1.6224)
= 2.56 * 10^4 kg ( 1.6224)
=4.153 * 10^4 J
As per guide lines I worked first problem please post remaining questions in the next post
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