Problem#1
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| Fig,1 |
A student performs a ballistic pendulum experiment using an apparatus similar to that shown in Figure 1B. She obtains the following average data: h = 8.68 cm, m1 = 68.8 g, and m2 = 263 g. The symbols refer to the quantities in Figure 1a. (a) Determine the initial speed v1A of the projectile. (b) The second part of her experiment is to obtain v1A by firing the same projectile horizontally (with the pendulum removed from the path), by measuring its final horizontal position x and distance of fall y (Fig. 2). Show that the initial speed of the projectile is related to x and y through the relation
v1A = x/(2y/g)1/2
What numerical value does she obtain for v1A based on her measured values of x = 257 cm and y = 85.3 cm? What factors might account for the difference in this value compared to that obtained in part (a)?
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| Fig.2 |
Answer:
(a) Utilizing conservation of momentum,
m1v1A = (m1 + m2)vB
where vB = (2gh)1/2
68.8gv1A = (68.8g + 263g)[2(9.80 m/s2)(8.68 x 10-2m)2]1/2
v1A = 6.29 m/s
(b) Utilizing the two equations,
y = ½ gt2 and x = v1At
we combine them to find
v1A = x/(2y/g)1/2
v1A = 2.57m/([2(0.853 m)/(9.80 m/s2)])1/2 = 6.16 m/s
Most of the 2% difference between the values for speed is accounted for by the uncertainty in the data, estimated as
0.01/8.68 + 0.1/68.8 + 1/263 + 1/257 + 0.1/85.3 = 1.1%
Problem#2
Small ice cubes, each of mass 5.00 g, slide down a frictionless track in a steady stream, as shown in Figure 3. Starting from rest, each cube moves down through a net vertical distance of 1.50 m and leaves the bottom end of the track at an angle of 40.0° above the horizontal. At the highest point of its subsequent trajectory, the cube strikes a vertical wall and rebounds with half the speed it had upon impact. If 10.0 cubes strike the wall per second, what average force is exerted on the wall?
Answer:
The ice cubes leave the track with speed determined by
mgyi = ½ mv2
(9.80 m/s2)(1.50 m) = ½ v2
v = 5.42 m/s
Its speed at the apex of its trajectory is
vi = 5.42 m/s cos40.00 = 4.15 m/s
For its collision with the wall we have
F∆t = mvf – mvi
F∆t = 0.005kg(–½ 4.15 m/s) – (0.005kg)(4.15 m/s)
F∆t = –3.12 x 10-2 Ns
The impulse exerted by the cube on the wall is to the right, +3.12 x 10-2kg.m/s. Here F could refer
to a large force over a short contact time. It can also refer to the average force if we interpret ∆t as
1/10 s, the time between one cube’s tap and the next’s.
to a large force over a short contact time. It can also refer to the average force if we interpret ∆t as
1/10 s, the time between one cube’s tap and the next’s.
Favr = +3.12 x 10-2 Ns/0.1s = 0.312 N to the right
Problem#3
A 5.00-g bullet moving with an initial speed of 400 m/s is fired into and passes through a 1.00-kg block, as in Figure 4. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring with force constant 900 N/m. If the block moves 5.00 cm to the right after impact, find (a) the speed at which the bullet emerges from the block and (b) the mechanical energy converted into
internal energy in the collision.
internal energy in the collision.
Answer:
(a) Find the speed when the bullet emerges from the block by using momentum conservation:
mvi = MVi + mv
The block moves a distance of 5.00 cm. Assume for an approximation that the block quickly reaches its maximum velocity, Vi , and the bullet kept going with a constant velocity, v. The block then
compresses the spring and stops.
compresses the spring and stops.
½ MVi2 = ½ kx2
(1.00kg)Vi2 = (900 N/m)(5.00 x 10-2 m)2
Vi = 1.50 m/s
Then from (1) we get
(5.00 x 10-3 kg)(400 m/s) = (1.00kg)(1.50m/s) + (5.00 x 10-3 kg)v
v = 100 m/s
(b) the mechanical energy converted into internal energy in the collision.
∆E = ∆U + ∆K
∆E = ½ kxf2 – 0 + ½ mvf2 – ½ mvi2
∆E = ½ (900 N/m)(5.00 x 10-2 m)2 + ½ (5.00 x 10-2 kg)(100 m/s)2 – ½ (5.00 x 10-2kg)(400 m/s)2
∆E = –374 J
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