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Car B is following Car A and has a greater speed than Car A. The two cars are moving in a straight line and in the same direction, and have the same mass. In situation one, Car A is traveling at 10 mph and Car B at 20 mph. In situation two, Car A is traveling at 30 mph and Car B at 40 mph. Assuming a perfectly inelastic collision in which the cars stick together after the collision, which of the following will be true?

a. The force of the collision in the two situations will be equal.
b. Situation two will cause the greater force of collision.
c. Situation one will cause the greater force of collision.

User Weng
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1 Answer

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Answer:

we see that force of collision is the same in both the case.

Step-by-step explanation:

Let the time of impact in both the situation be t and mass of each be m

Applying conservation of momentum in the first case

m₁v₁ + m₂v₂ = (m₁ +m₂ ) v

m x 20 + m x 10 = 2m x v

v = 15 mph.

So the speed of B will be reduced from 20 to 15 mph and speed of A will be increased from 10 to 15 mph.

Considering impact on B only

Impulse on B is equal to change in momentum

F X t = m ( 20 - 15 )

F is force of collision .

F = 5m / t

In the second case ,

Applying conservation of momentum in the second case

m₁v₁ + m₂v₂ = (m₁ +m₂ ) v

m x 40 + m x 30 = 2m x v

v = 35 mph.

So the speed of B will be reduced and speed of A will be increased.

Considering impact on B only

Impulse on B is equal to change in momentum

F X t = m ( 40 - 35 )

F is force of collision .

F = 5m / t

So we see that force of collision is the same in both the case.

User Satya Chandra
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