Problem#1
As shown in Figure 1, 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?
Answer:
Momentum of the bob-bullet system is conserved in the collision:
mv = m(v/2) + Mu
m(v/2) = Mu (*)
with u is bob velocity after the impact.
Energy is conserved for the bob-Earth system between bottom and top of swing. At the top the stiff rod is in compression and the bob nearly at rest.
Ki + Ui = Kf + Uf
½ Mu2 + 0 = 0 + Mg(2l)
u = (4gl)1/2
then from (*)
m(v/2) = M(4gl)1/2
v = 4M(gl)1/2/m
Problem#2
A 12.0-g wad of sticky clay is hurled horizontally at a 100-g wooden block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides 7.50 m before coming to rest. If the coefficient of friction between the block and the surface is 0.650, what was the speed of the clay immediately before impact?
Answer:
After impact, the change in kinetic energy of the clay-block-surface system is equal to the increase in internal energy:
∆Emech = ∆K + ∆U
–fd = ½ (m1 + m2)v22 – 0 + 0 + 0
ยต(m1 + m2)gd = ½ (m1 + m2)v22
(0.650)(9.80 m/s2)(7.50 m) = ½ v22
v2 = 9.77 m/s
At impact, momentum of the clay-block system is conserved, so:
m1v1 = (m1 + m2)v2
(12.0 x 10-3 kg)v1 = (12.0 x 10-3 kg + 0.100 kg)(9.77 m/s)
v1 = 91.2 m/s
Problem#3
A 7.00-g bullet, when fired from a gun into a 1.00-kg block of wood held in a vise, penetrates the block to a depth of 8.00 cm. What If? This block of wood is placed on a frictionless horizontal surface, and a second 7.00-g bullet is fired from the gun into the block. To what depth will the bullet penetrate the block in this case?
Answer:
We assume equal firing speeds v and equal forces F required for the two bullets to push wood fibers
apart. These equal forces act backward on the two bullets.
For the first,
∆Emech = ∆K = Kf – Ki
–F(∆x) = 0 – ½ mvi2
F(0.080 m) = ½ (7.00 x 10-3 kg)v2
F = 4.375 x 10-2v2 (1)
For the second,
pi = pf
(7.00 x 10-3 kg)v = (1.014 kg)vf
vf = 6.90 x 10-3v (2)
we use again
∆Emech = Kf – Ki
–Fd = ½ (1.014kg)vf2 – ½ (7.00 x 10-3 kg)v2
Substituting (1) and (2)
–(4.375 x 10-2v2)d = (0.507kg)(6.90 x 10-3v)2 – (3.50 x 10-3 kg)v2
–(4.375 x 10-2)d = (0.507kg)(6.90 x 10-3)2 – (3.50 x 10-3 kg)
–(4.375 x 10-2)d = (0.507kg)(6.90 x 10-3)2 – (3.50 x 10-3 kg)
–(4.375 x 10-2)d = –3.47 x 10-3 m
d = 0.0794 m = 7.94 cm
Problem#4
(a) Three carts of masses 4.00 kg, 10.0 kg, and 3.00 kg move on a frictionless horizontal track with speeds of 5.00 m/s, 3.00 m/s, and 4.00 m/s, as shown in Figure 3. Velcro couplers make the carts stick together after colliding. Find the final velocity of the train of three carts. (b) What If?
Does your answer require that all the carts collide and stick together at the same time? What if they collide in a different order?
Answer:
(a) Using conservation of momentum
pf = pi
(4.00 kg + 10.0kg + 3.00 kg)vf = (4.00 kg)(5.00 m/s) + (10.0 kg)(3.00 m/s) + (3.00kg)(–4.00 m/s)
vf = 2.24 m/s toward the right
(b) No. For example, if the 10-kg and 3.0-kg mass were to stick together first, they would move with a speed given by solving
(4.00 kg + 10.0kg + 3.00 kg)v = (10.0 kg)(3.00 m/s) + (3.00kg)(–4.00 m/s)
v = 1.38 m/s
Then when this 13 kg combined mass collides with the 4.0 kg mass, we have
(17.0 kg)v = (13.0 kg)(1.38 m/s) + (4.00kg)(5.00 m/s)
v = 2.24 m/s
just as in part (a). Coupling order makes no difference.
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