PHYSICS 113 – SPRING 2018 2nd in-class MIDTERM EXAM
INSTRUCTIONS
Answer all questions on the exam. Partial credit can only be given for work shown. Note that credit will not be given for just the answer, even if correct, and also that credit may be taken away for a correct answer preceded by incorrect work. If any clarification is needed concerning the questions or what is being asked, please ask me. Remember to clearly specify vectors and units for all phys- ical quantities. GOOD LUCK!
Problem A: People jumping off a flatcar
Two people, each of mass mp, stand on a railway flatcar of mass mc. They jump off one end of the flatcar with velocity u relative to the car. The car rolls in the opposite direction without friction. (i) What is the final velocity of the car if the two people jump off at the same time? (ii) What is the final velocity of the car if the people jump off one at a time? (iii) Which of these velocities is larger in magnitude?
Problem B: Recoil
A small cube of mass m1 slides down a circular track of radius R cut into a large block of mass m2 as shown in Fig.1. The large block rests on a table, and both blocks move without friction. The blocks are initially at rest, and m1 starts from the top of the path. Find the velocity v⃗ of the cube as it leaves the block.
Problem C: Particles colliding on inclined plane
A small sphere of mass m = 5kg starts with an initial velocity of v = 60m/s. It will collide and stick to a second small sphere of mass M = 10Kg. The two spheres will slide up a frictionless inclined plane of angle α = 35X, See Fig.2. What is the velocity of the two spheres after the collision? Assuming the radii of the spheres are negligible, what is the maximum height reached by the two spheres?
1
Figure 1
m
vM
Figure 2
α
Solutions Problem A
Choose a reference system at rest with the ground. The total initial momentum of the car plus people is zero and this is always conserved as the system is isolated. The only momentum that matters here is that in the x direction. (i) Consider the case of the two people jumping off at the same time. Be vc the velocity of the car as the to people jump. Conservation of momentum gives
0 = mcvc + 2mpvp (1) The velocity of the people vp as they jump off the car in the ground reference
frame is given by vp = vc + u and therefore
0=mcvc +2mp(vc +u) (2)
from which we find
(ii) Now let us solve the case of the people jumping off one at a time. Following the same reasoning as above, after the first person jumps off, conservation of momentum will give
0 = (mc + mp)vc,1 + mp(vc,1 + u) (4) which yields a velocity of the car
−mpu vc
vc,1=m+2m =2 (5)
cp
Now let us consider the second jump. At this point the total momentum of the car plus one person is given by
p1 = (mc + mp)vc,1 (6) As the second person jumps momentum is conserved and we have
(mc + mp)vc,1 = mcvc,2 + mp(vc,2 + u) (7) which gives a velocity of the car
(mc + mp)vc,1 − mpu
vc,2 = m+m (8)
cp mpu
= vc,1−m+m (9) cp
11
= −mpu(m +2m +m +m ) (10)
cpcp
(iii) Notice that ∣vc,1∣ < ∣mpu/(mc + mp)∣ and therefore ∣vc,2∣ > vc.
2
−2mp u
vc = m + 2m (3)
cp
Problem B
In the system block-cube there are no external forces in the x−direction so mo- ment is conserved in that direction, px = const. Before the cube starts sliding down, the total momentum is zero, and so it must be zero also at the final moment when the cube leaves the block:
0 = m1v1f + m2v2f (11) so the final velocity of the block is
m1
v2f =−m v1f (12)
2
Conservation of energy between the instants i and f gives 1212
m1gR = 2m1v1f + 2m2v2f (13) as we put the zero of potential energy at the height at which the cube leaves the
block. Using the expression for v2f in the above equation we find
1 2 1 m1 2 m1gR=2m1v1f+2m2(m v1f) (14)
and solving for v1f we finally find
√
v1f =
Problem C
In this problem, momentum is conserved in the x−direction. This helps finding the common velocity vf of the two blocks after the collision.
mv
mv=(m+M)vf →vf = m+M =20m/s (16)
The two spheres stick together so the collision is inelastic and some kinetic energy is lost. After collision, though, energy is conserved, and we can use that to find the maximum height of the spheres, reached when they come to a stop:
12
2
2gRm2 m +m
(15)
from which we get
12
2(m+M)vf =(m+M)gh (17)
vf2 2g
3
h =
= 20.38m (18)