Question

Two 100 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car B at −10 m/s when they collide head–on. If the resulting velocity of Car B after the collision is 8 m/s, what is the velocity of Car A after the collision?

Answer 1

Answer:

The velocity of Car A after the collision is 10 [m/s] in opposite direction to its original movement

Explanation:

In order to solve this problem we must use the principle of conservation of linear momentum, which is defined as the product of mass by Velocity.

P = m*v

where:

P = momentum [kg*m/s]

m = mass [kg]

v = velocity.

We have to study the moment before the collision and the moment after the collision, the moment before the collision will be taken to the left of the equal sign and the moment after the collision will be taken to the right of the equal sign.

ΣPbefore = ΣPafter

(m1*v1) + (m2*v2) = (m1*v3) + (m2*v4)

where:

m1 = m2 = 100 [kg]

V1 = velocity of car A = 8 [m/s]

V2 = velocity of car B = - 10 [m/s]

V4 = velocity of car B after the collision = 8 [m/s]

(100*8) - (100*10) = (100*V3) + (8*100)

V3 = -10 [m/s]

Step-by-step explanation:

Answer 2

The velocity of Car A after the collision is 10 [m/s] in opposite direction to its original movement

Step-by-step explanation:

In order to solve this problem we must use the principle of conservation of linear momentum, which is defined as the product of mass by Velocity.

P = m*v

where:

P = momentum [kg*m/s]

m = mass [kg]

v = velocity.

We have to study the moment before the collision and the moment after the collision, the moment before the collision will be taken to the left of the equal sign and the moment after the collision will be taken to the right of the equal sign.

ΣPbefore = ΣPafter

(m1*v1) + (m2*v2) = (m1*v3) + (m2*v4)

where:

m1 = m2 = 100 [kg]

V1 = velocity of car A = 8 [m/s]

V2 = velocity of car B = - 10 [m/s]

V4 = velocity of car B after the collision = 8 [m/s]

(100*8) - (100*10) = (100*V3) + (8*100)

V3 = -10 [m/s]