Collisions in One Dimension Problems and Solutions 1

 Problem#1

Two blocks are free to slide along the frictionless wooden track ABC shown in Figure 1. The block of mass m₁ = 5.00 kg is released from A. Protruding from its front end is the north pole of a strong magnet, repelling the north pole of an identical magnet embedded in the back end of the block of mass m₂ = 10.0 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision.

Answer:
By using the law of conservation of mechanical energy, we will get the velocity of m₁ before collision

½ m₁v₁² = m₁gh

with v₁ = speed of m1at B before collision.

v₁² = 2(9.80 m/s²)(5.00 m)

v₁ = 9.90 m/s

We use the law of conservation of momentum,

m₁v₁ + 0 = m₁v_1f + m₂v_2f

(5.00 kg)(9.90 m/s) = (5.00 kg)v_if + (10.0 kg)v_2f

v_1f, speed of m₁ at B just after collision.

Because the collision is perfectly it applies

v_2f – v_1f = v_1i – v_2i

v_2f – v_if = 9.90 m/s – 0 or

v_2f = 9.90 + v_1f

then,
(5.00 kg)(9.90 m/s) = (5.00 kg)v_if + (10.0 kg)(9.90 + v_1f)

–49.5 kgm/s = (15.0 kg)v_1f

v_1f = –3.30 m/s

At the highest point (after collision)

m₁gy_max = ½mv_1f²

(–3.30 m/s)² = 2(9.80 m/s²)y_max

y_max = 55.6 cm

Problem#2
A 45.0-kg girl is standing on a plank that has a mass of 150 kg. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless supporting surface. The girl begins to walk along the plank at a constant speed of 1.50 m/s relative to the plank. (a) What is her speed relative to the ice surface? (b) What is the speed of the plank relative to the ice surface?

Answer:

Let v_g and v_p be the velocity of the girl and the plank relative to the ice surface. Then we may say that v_g – v_p is the velocity of the girl relative to the plank, so that

v_g – v_p = 1.50 m/s            (*)

Since total momentum of the girl-plank system is zero relative to the ice surface. Therefore
m_gv_g + m_pv_p = 0

45.0v_g + 150v_p = 0

v_g = –3.33v_p

Putting this into the equation (*) above gives

–3.33v_p – v_p = 1.50 m/s

v_p = –0.346 m/s and

v_g = –3.33(–0.346 m/s) = 1.15 m/s

Problem#3
Most of us know intuitively that in a head-on collision between a large dump truck and a  subcompact car, you are better off being in the truck than in the car. Why is this? Many people imagine that the collision force exerted on the car is much greater than that experienced by the truck. To substantiate this view, they point out that the car is crushed, whereas the truck is only dented. This idea of unequal forces, of course, is false. Newton’s third law tells us that both objects experience forces of the same magnitude. The truck suffers less damage because it is made of stronger metal. But what about the two drivers? Do they experience the same forces? To answer this question, suppose that each vehicle is initially moving at 8.00 m/s and that they undergo a perfectly inelastic head-on collision. Each driver has mass 80.0 kg. Including the drivers, the total vehicle masses are 800 kg for the car and 4 000 kg for the truck. If the collision time is 0.120 s, what force does the seatbelt exert on each driver?

Answer:
For the car-truck-driver-driver system, momentum is conserved:

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

(4000 kg)(8.00 m/s)i + (800 kg)(8.00 m/s)(–i) = (4000 kg + 800 kg)v_fi

v_f = (25600 kg.m/s)/(4800 kg) = 5.33 m/s

For the driver of the truck, the impulse-momentum theorem is

F∆t = m∆v

F(0.120 s) = (80 kg)[(5.33 m/s)i – (8.00 m/s)i]

F = –1.78kNi on the truck driver

For the driver of the car,

F(0.120 s) = (80 kg)[(5.33 m/s)i – (8.00 m/s)(–i)]

F = 8.89 kNi on the car driver

Problem#4
A neutron in a nuclear reactor makes an elastic headon collision with the nucleus of a carbon atom initially at rest. (a) What fraction of the neutron’s kinetic energy is transferred to the carbon  nucleus? (b) If the initial kinetic energy of the neutron is 1.60 x 10^-13 J, find its final kinetic energy and the kinetic energy of the carbon nucleus after the collision. (The mass of the carbon nucleus is nearly 12.0 times the mass of the neutron.)

Answer:
M = mass of carbon and m = mass of neutron

(a) the fraction of total kinetic energy transferred to the moderator is

K_f,carbon/K_i,neutron = f = ½ Mv_cf²/(½mv_ni²)

but M = 12m, then

f = 12v_cf²/v_ni²      (1)

momentum is conserved is

mv_ni = mv_nf + Mv_cf

v_ni – v_nf = (M/m)v_cf = 12v_cf

v_ni – v_nf = 12v_cf      (2)

Kinetic energy is conserved is

½ mv_ni² = ½ mv_nf² + ½ Mv_cf²

v_ni² – v_nf² = (M/m)v_cf²

v_ni² – v_nf² = 12v_cf²              (3)

divide (3) and (2) we get

v_ni + v_nf = v_cf        (4)

add (2) and (4), we get

2v_ni = 13v_cf or v_cf = 2v_ni/13, then

f = 12(2/13)² = 48/169 or 28.4%

(b) final kinetic energy and the kinetic energy of the carbon nucleus after the collision is

K_f,carbon/K_i,neutron = 28.4%

K_f,carbon = (0.284)K_i,neutron = (0.284)(1.60 x 10^-13 J) = 4.54 x 10^-14 J and

K_i,neutron = (0.716)(1.60 x 10^-13 J) = 1.15 x 10^-13 J   

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