Problem#1
Two gliders are set in motion on an air track. A spring of force constant k is attached to the near side of one glider. The first glider, of mass m1, has velocity v1, and the second glider, of mass m2, moves more slowly, with velocity v2, as in Figure 1. When m1 collides with the spring attached to
m2 and compresses the spring to its maximum compression x max, the velocity of the gliders is v. In terms of v1, v2, m1, m2, and k, find (a) the velocity v at maximum compression, (b) the maximum compression xmax, and (c) the velocity of each glider after m1 has lost contact with the spring.
Answer:
Conservation of momentum:
m1v1i + m2v2i = m1v1f + m2v2f or
m1(v1i – v1f) = m2(v2f – v2i) (1)
Conservation of kinetic energy:
½m1v1i2 + ½m2v2i2 = ½m1v1f2 + ½m2v2f2 or
m1(v1i2 – v1f2) = m2(v2f2 – v2i2) (2)
from (1) and (2) we get
(v1i2 – v1f2)/(v1i – v1f) = (v2f2 – v2i2)/(v2f – v2i)
(v1i + v1f) = (v2f + v2i) or
v2f = v1i + v1f – v2i
Substituting equation (3) into equation (1):
m1(v1i – v1f) = m2(v1i + v1f – v2i – v2i)
(m1 + m2)v1f = (m1 – m2)v1i + 2m2v2i
v1f = (m1 – m2)v1i/(m1 + m2) + 2m2v2i/(m1 + m2)
and
v2f = (m2 – m1)v2i/(m1 + m2) + 2m1v1i/(m1 + m2)
Problem#2
Review problem. A 60.0-kg person running at an initial speed of 4.00 m/s jumps onto a 120-kg cart initially at rest (Figure 2). The person slides on the cart’s top surface and finally comes to rest relative to the cart. The coefficient of kinetic friction between the person and the cart is 0.400.
Friction between the cart and ground can be neglected. (a) Find the final velocity of the person and cart relative to the ground. (b) Find the friction force acting on the person while he is sliding across the top surface of the cart. (c) How long does the friction force act on the person? (d) Find the
change in momentum of the person and the change in momentum of the cart. (e) Determine the displacement of the person relative to the ground while he is sliding on the cart. (f) Determine the displacement of the cart relative to the ground while the person is sliding. (g) Find the change in kinetic energy of the person. (h) Find the change in kinetic energy of the cart. (i) Explain why the answers to (g) and (h) differ. (What kind of collision is this, and what accounts for the loss of mechanical energy?)
Answer:
(a) the final velocity of the person and cart relative to the ground, conservation of momentum gives
mpvpi + mcvci = mpvpf + mcvcf = (mp + mc)vf
(60.0kg)(4.00 m/s) + 0 = (60.0 kg + 120kg)vf
vf = (1.33m/s)i
(b) the friction force acting on the person while he is sliding across the top surface of the cart is
fk = µkn = 0.400(6.00kg)(9.80 m/s2) = 235N
or fk = 235Ni
(c) For the person,
I = pf – pi
–ft = mp(vpf – vpi)
–235Nt = (60.0kg)(1.33m/s – 4.00 m/s)
t = 0.680s
(d) the change in momentum of the person is
∆p = mp(vpf – vpi) = (60.0kg)(1.33i m/s – 4.00i m/s) = –160i kg.m/s
and the change in momentum of the cart is
∆pc = mc(vcf – vci) = (120kg)(1.33i m/s – 0) = +160i kg.m/s
(e) the displacement of the person relative to the ground while he is sliding on the cart is
∆xp = ½ (vpi + vpf)t = ½ (4.00 m/s + 1.33 m/s)(0.680s) = 1.81 m
(f) the displacement of the cart relative to the ground while the person is sliding is
∆xc = ½ (vci + vcf)t = ½ (0 + 1.33 m/s)(0.680s) = 0.454 m
(g) the change in kinetic energy of the person is
∆Kp = ½ mp(vpf2 – vpi2)
∆Kp = ½ (60.0kg)[(1.33 m/s)2 – (4.00m/s)2] = –427J
(h) the change in kinetic energy of the cart is
∆Kc = ½ mc(vcf2 – vci2)
∆Kc = ½ (120kg)[(1.33 m/s)2 – 0] = 107J
(i) The force exerted by the person on the cart must equal in magnitude and opposite in
direction to the force exerted by the cart on the person. The changes in momentum of
the two objects must be equal in magnitude and must add to zero. Their changes in
kinetic energy are different in magnitude and do not add to zero. The following
from the distance moved by the point of application of the friction force to the cart.
The total change in mechanical energy for both objects together, J, becomes
+320 J of additional internal energy in this perfectly inelastic collision.
Problem#3
A golf ball (m = 46.0 g) is struck with a force that makes an angle of 45.0° with the horizontal. The ball lands 200 m away on a flat fairway. If the golf club and ball are in contact for 7.00 ms, what is the average force of impact? (Neglect air resistance.)
Answer:
The equation for the horizontal range of a projectile is
R = vi2 sin2α/g
Then the initial velocity is
200m = vi2 sin(2 x 450)/(9.80 m.s-2)
vi = 44.3 m/s
the impulse is the same as the change in momentum of the ball,
I = ∆p
Favr.∆t = mb(vi – 0)
Therefore, the magnitude of the average force acting on the ball during the impact is:
Favr = mbvi/∆t
Favr = (4.60 x 10-2kg)(44.3 m/s)/(7.00 x 10-3 s)
Favr = 291 N
Problem#4
An 80.0-kg astronaut is working on the engines of his ship, which is drifting through space with a constant velocity. The astronaut, wishing to get a better view of the Universe, pushes against the ship and much later finds himself 30.0 m behind the ship. Without a thruster, the only way to return to the ship is to throw his 0.500-kg wrench directly away from the ship. If he throws the wrench with a speed of 20.0 m/s relative to the ship, how long does it take the astronaut to reach the ship?
Answer:
We hope the momentum of the wrench provides enough recoil so that the astronaut can reach the
ship before he loses life support! We might expect the elapsed time to be on the order of several
minutes based on the description of the situation. No external force acts on the system (astronaut plus wrench), so the total momentum is constant. Since the final momentum (wrench plus astronaut) must be zero, we have final momentum = initial momentum = 0. Then,
mwvw + mava = 0
thus, va = –mwvw/ma = –(0.500kg)(20.0m/s)/(80.0kg) = –0.125 m/s
At this speed, the time to travel to the ship is
t = x/va = 30.0m/(0.125 m/s) = 240s = 4.00 minutes
The astronaut is fortunate that the wrench gave him sufficient momentum to return to the ship in a
reasonable amount of time! In this problem, we were told that the astronaut was not drifting away
from the ship when he threw the wrench. However, this is not quite possible since he did not
encounter an external force that would reduce his velocity away from the ship (there is no air
friction beyond earth’s atmosphere). If this were a real-life situation, the astronaut would have to
throw the wrench hard enough to overcome his momentum caused by his original push away from
the ship.
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