Inelastic Collision Problems and Solutions

 Problem #1

A 1850 kg luxury sedan stopped at a traffic light is struck from the rear by a compact car with a mass of 975 kg. The two cars become entangled as a result of the collision. If the compact car was moving at a velocity of 22.0 m/s to the north before the collision, what is the velocity of the entangled mass after the collision?

Answer:
Given: m₁ = 1850 kg, m₂ = 975 kg, v_1,i = 0 m/s, v_2,i = 22.0 m/s to the north
Use the equation for a perfectly inelastic collision.

m₁v_1,i + m₂v_2,i = (m₁ + m₂)v_f

v_f = [m₁v_1,i + m₂v_2,i]/(m₁ + m₂)

v_f = [(1850 kg)(0 m/s) + (975 kg)(22.0 m/s north)]/(1850 kg + 975 kg)

v_f = 7.59 m/s to the north

Problem #2
Two clay balls collide head-on in a perfectly inelastic collision. The first ball has a mass of 0.500 kg and an initial velocity of 4.00 m/s to the right. The second ball has a mass of 0.250 kg and an initial velocity of 3.00 m/s to the left. What is the decrease in kinetic energy during the collision?

Answer:
Given: m₁ = 0.500 kg, m₂ = 0.250 kg, v_1,i = 4.00 m/s to the right, v_1,i = +4.00 m/s, v_2,i = 3.00 m/s to the left, v_2,i = −3.00 m/s

Choose an equation or situation: The change in kinetic energy is simply the initial kinetic energy subtracted from the final kinetic energy

∆K = K_f − K_i

Determine both the initial and final kinetic energy

Initial: K_i = K_1,i + K_2,i = ½m₁v_1,i ² + ½m₂v_2,i²
Final : K_f = K_1,f + K_2,f = ½(m₁ + m₂) v_f²

As you did in Sample Problem E, use the equation for a perfectly inelastic collision to calculate the final velocity.
Substitute the values into the equation and solve: First, calculate the final velocity, which will be used in the final kinetic energy equation.

v_f = [(0.500 kg)(4.00 m/s) + (0.250 kg)(−3.00 m/s)]/(0.500 kg + 0.250 kg)
v_f = 1.67 m/s to the right

Next calculate the initial and final kinetic energy.

K_i = ½ (0.500 kg)(4.00 m/s)² + ½ (0.250 kg)(−3.00 m/s)² = 5.12 J
K_f = ½ (0.500 kg + 0.250 kg)(1.67 m/s)² = 1.05 J

Finally, calculate the change in kinetic energy.

∆K = K_f − K_i = 1.05 J − 5.12 J
∆K = −4.07 J

The negative sign indicates that kinetic energy is lost.

Problem #3

Fig.1

As shown in Fig. 1, a bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed v/2. The pendulum bob is suspended by a stiff rod of length ` and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle?
Answer:
Whoa! There’s a hell of a lot of things going on in this problem. Let’s try to sort them out. We break things down into a sequence of events: First, the bullet has a very rapid, very strong interaction with the pendulum bob, where is quickly passes through, imparting a velocity to the bob which at first will have a horizontal motion. Secondly, the bob swings upward and, as we are told, will get up to the top of the vertical circle.
We show the collision in Fig. 3. In this rapid interaction there are no net external forces acting on the system that we need to worry about. So its total momentum will be conserved. The total horizontal momentum before the collision is

P_i,x = mv + 0 = mv

If after the collision the bob has velocity v’, then the total momentum is

P_f,x = m(v/2) + Mv’

Conservation of momentum, P_i,x = P_f,x gives

mv = m(v/2) + Mv’
Mv’ = m(v/2

and so:
v’ = mv/2M                                         (7.14)

Now consider the trip of the pendulum bob up to the top of the circle (it must get to the top, by assumption). There are no friction–type forces acting on the system as M moves, so mechanical energy is conserved.
If we measure height from the bottom of the swing, then the initial potential energy is zero while the initial kinetic energy is

K_i = ½ M(v’)²

Fig.2

  
Now suppose at the top of the swing mass M has speed v_top. Its height is 2l and its potential energy is Mg(2l) so that its final energy is

E_f = ½ Mv²_top + 2Mgl

so that conservation of energy gives:

½ M(v’)² = ½ Mv²_top + 2Mgl                                          (7.15)

What do we know about v_top? A drawing of the forces acting on M at the top of the swing is shown in Fig. 7.6. Gravity pulls down with a force Mg. There may be a force from the suspending rod; here, I’ve happened to draw it pointing upward. Can this force point upward? Yes it can. . .we need to read the problem carefully. It said the bob was suspended by a stiff rod and such an object can exert a force (still called the tension T) inward or outward along its length. (A string can only pull inward.) The bob is moving on a circular path with (instantaneous) speed vtop so the net force on it points downward andhas magnitude Mv²_top/l`:

Mg − T = Mv²_top
.
Since T can be positive or negative, vtop can take on any value. It could be zero. What condition are we looking for which corresponds to the smallest value of the bullet speed v?
We note that as v gets bigger, so does v’ (the bob’s initial speed). As v’ increases, so does v_top, as we see from conservation of energy. But it is entirely possible for vtop to be zero, and that will give the smallest possible value of v. That would correspond to the case where M picked up enough speed to just barely make it to the top of the swing. (And when the bob goes past the top point then gravity moves it along through the full swing.)

So with v_top = 0 then Eq. 7.15 gives us

½ M(v’)² = 2Mgl
v'² = 4gl

and:

v’ = [4gl]^1/2

and putting this result back into Eq. 7.15, we have

[4gl]^1/2 = mv/2M

Finally, solve for v:

v = [2M/m][4gl]^1/2
v = 4M[gl]^1/2/m

The minimum value of v required to do the job is v = 4M[gl]^1/2/m

Problem #4
In a curling match, a 6.0­kg rock with speed 3.50 m/s collides with another motionless 6.0­kg rock. What are the velocities of the rocks after the collision if it is (a) elastic or (b) totally inelastic? (c) How much energy was lost in the inelastic collision? Ignore friction and assume all motion is in a straight line.

Answer:
(a) In an elastic collision, both momentum and kinetic energy is conserved. Thus we have the equations;

m₁v_1f + m₂v_2f = m₁v_1i + m₂v_2i , (1)
v_1f  +­ v_2f = ­(v_1i +­ v_2i ) . (2)

Since v_2i = 0, the two equations can be combine to yield

v_1f = (m₁ – m₂)v_1i/(m₁ + m₂) = 0

And
v_2f = (2m₂)v_1i/(m₁ + m₂) = 2 x 6 kg x 350 m/s/(6 kg + 6 kg) = 350 m/s

So after the collision, the first rock comes to a complete halt and the second rock takes off with the velocity of the first rock before the collision.

(b) In a totally inelastic collision, the two rocks stick together so that conservation of momentum becomes

(m₁ + m₂ )v_f = m₁v_1i + m₂v_2i .

since v_2i = 0,

v_f = m₁v_1i/(m₁ + m₂) = (6 kg x 3.5 m/s)/(6 kg + 6 kg) = 1.75 m/s .

(c) We find the kinetic energy lost in the inelastic collision by examining the energies just before and after the collision:

K_i = ½m₁(v_1i)² + ½m₂(v_2i)² = ½(6 kg)(3.5 m/s)² + 0 = 36.75 J ,
K_f = ½(m₁+m₂)(v_f)² = ½(6 kg + 6 kg)(1.75 m/s)² = 18.375 J .

So the change in energy is E = K_f ­– K_i = ­18.4 J. Thus 18.4 J of energy was lost in the collision.   

Share :