Conservation of Linear Momentum Problems and Solutions

 Problem #1

A 10.0g bullet is stopped in a block of wood (m = 5.00 kg). The speed of the bullet–plus–wood combination immediately after the collision is 0.600 m/s. What was the original speed of the bullet?

Fig.1

Answer:
A picture of the collision just before and after the bullet (quickly) embeds itself in the wood is given in Fig. 1. The bullet has some initial speed v0 (we don’t know what it is.) The collision (and embedding of the bullet) takes place very rapidly; for that brief time the bullet and block essentially form an isolated system because any external forces (say, from friction from the surface) will be of no importance compared to the enormous forces between the bullet and the block. So the total momentum of the system will be conserved; it is the same before and after the collision.
In this problem there is only motion along the x axis, so we only need the condition that the total x momentum (P_x) is conserved.
Just before the collision, only the bullet (with mass m) is in motion and its x velocity is v₀. So the initial momentum is

P_i,x = mv₀ = (10.0 × 10⁻³ kg)v₀

Just after the collision, the bullet–block combination, with its mass of M + m has an x velocity of 0.600 m/s . So the final momentum is

P_f,x = (M + m)v = (5.00 kg + 10.0 × 10⁻³ kg)(0.600 m/s) = 3.01 kg·m/s

Since P_i,x = P_f,x, we get:

(10.0 × 10⁻³ kg)v₀ = 3.01 kg·m/s
v₀ = 301 m/s

The initial speed of the bullet was 301 m/s.

Problem #2
While chasing an armed suspect into and onto an ice rink, a police constable is shot. Fortunately, the constable is wearing a bullet­proof vest which absorbs the bullet. If the muzzle velocity of the bullet is 350 m/s and the its mass is 100 g. Find the final velocity of the constable and bullet if her mass is 69.5 kg. Assume all motion is in a straight line and ignore friction. Assume that the constable is at rest.

Fig.2

Answer:
We have a totally inelastic collision, so momentum is conserved. For this particular problem

(m_police + m_bullet)v_pf = m_policev_pi + m_bulletv_bi

Since we are told v_pi = 0,

v_pf = m_bulletv_bi/(m_police + m_bullet)
v_pf = (0.100 kg)(­350 m/s)/(69.5 kg + 0.1 kg) = ­0.503 m/s .

So the constable is knocked backwards at 0.50 m/s.


Problem #3
Two 22.7kg ice sleds are placed a short distance apart, one directly behind the other, as shown in Fig. 3 (a). A 3.63 kg cat, standing on one sled, jumps across to the other and immediately back to the first. Both jumps are made at a speed on 3.05 m/s relative to the ice. Find the final speeds of the two sleds.

Fig.3

Answer
We will let the x axis point to the right. In the initial picture (not shown) the cat is sitting on the left sled and both are motionless. Taking our system of interacting “particles” to be the cat and the left sled, the initial momentum of the system is P = 0.
After the cat has made its first jump, the velocity of this sled will be v_L,x, and the (final) total momentum of the system will be

P_f = (22.7kg)v_L,x + (3.63 kg)(+3.05 m/s)

Note, we are using velocities with respect to the ice, and that is how we were given the velocity of the cat. Now as there are no net external forces, the momentum of this system is conserved. This gives us:

0 = (22.7kg)v_L,x + (3.63 kg)(+3.05 m/s)

with which we easily solve for v_L,x:

v_L,x = −(3.63 kg)(+3.05 m/s)/(22.7kg) = −0.488 m/s

so that the left sled moves at a speed of 0.488 m/s to the left after the cat’s first jump.

The cat lands on the right sled and after landing it moves with the same velocity as that sled; the collision here is completely inelastic. For this part of the problem, the system of “interacting particles” we consider is the cat and the right sled. (The left sled does not interact with this system.) The initial momentum of this system is just that of the cat,

P_i = (3.63 kg)(+3.05 m/s) = 11.1 kg·m/s

If the final velocity of both cat and sled is v_R,x then the final momentum is

P_f = (22.7kg +3.63 kg)v_R,x = (26.3kg)v_R,x

(The cat and sled move as one mass, so we can just add their individual masses.) Conservation of momentum of this system, P_i = P_f gives

11.1 kg·m/s = (26.3kg)v_R,x 

so

v_R,x = 11.1 kg·m/s/(26.3kg) = 0.422 m/s

Now we have the velocities of both sleds as they are pictured in Fig. 7 (b). And now the cat makes a jump back to the left sled, as shown in Fig. 7 (a). Again, we take the system to be the cat and the right sled. Its initial momentum is

P_i = (22.7kg +3.63 kg)(0.422 m/s) = 11.1 kg·m/s

Now after the cat leaps, the velocity of the cat (with respect to the ice) is −3.05 m/s , as specified in the problem. If the velocity of the right sled after the leap is v’_R,x then the final momentum of the system is

P_f = (3.63 kg)(−3.05 m/s) + (22.7kg)v’_R,x

Conservation of momentum for the system, P_i = P_f , gives

11.1 kg·m/s = (3.63 kg)(−3.05 m/s) + (22.7kg)v’_R,x

so that we can solve for v’_R,x:

v’_R,x = (11.1 kg·m/s + 11.1 kg·m/s)(22.7kg) = 0.975 m/s

so during its second leap the cat makes the right sled go faster!
Finally, for the cat’s landing on the left sled we consider the (isolated) system of the cat and the left sled. We already have the velocities of the cat and sled at this time; its initial momentum is

P_i = (22.7 kg)(−0.488 m/s) + (3.63 kg)(−3.05 m/s) = −22.1 kg·m/s

After the cat has landed on the sled, it is moving with the same velocity as the sled, which we will call v’_L,x. Then the final momentum of the system is

P_f = (22.7kg +3.63 kg)v’_L,x = (26.3kg)v’_L,x

And momentum conservation for this collision gives

−22.1 kg·m/s = (26.3kg)v’_L,x

and then

v’_L,x = (−22.1 kg·m/s)/(26.3kg) = −0.842 m/s

Summing up, the final velocities of the sleds (after the cat is done jumping) are:

Left Sled: v’_L,x = −0.842 m/s
Right Sled: v’_r,x = +0.975 m/s

Problem #4
A 60.0­kg person, running horizontally with a velocity of 3.80 m/s jumps on a 12.0­kg sled that is initially at rest. (a) Ignoring the effects ofstatic friction, find the velocity of the sled and person as they move away. (b) The sled and person coast 30.0 m on levelsnow before coming to a rest. What is the coefficient of kinetic friction between the sled and the snow?

Answer:
(a) We have a totally inelastic collision, so

 (m_person + m_sled)v_f = m_personv_person + m_sledv_sled

since v_sled = 0,

v_f = m_personv_person/(m_person + m_sled)
v_f = (60 kg x 3.8 m/s)/(60 kg + 12 kg) = 3.17 m/s

(b) This portion of the question involves a force and a distance suggesting the use of Work-­Energy methods. Since there is friction, W_NC = W_friction is not zero. We need a free body diagram to find f_k .

Using Newton's Second Law, the equation in the second column tells us that N = mg. Since f_k = μ_kN, we have f_k = μ_kmg. So the work done by friction is

W_friction = f_kΔx cos(180°) = ­μ_kmgΔx

As well, we know that

W_friction = E_f ­– E_i = ­½mv²

Combing these two results yields,

μ_kmgΔx = ­½mv²

Solving for μ_k

μ_k = ½v²/gΔx
μ_k = ½(3.17 m/s)²/(9.81 m/s² × 30 m) = 0.017

The coefficient of kinetic friction was 0.017.  

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