Two bumper cars in an amusement park ride collide elas- tically as one approaches the other directly from the rear (Fig. 7–43). Car A has a mass of 435 kg and car B 495 kg, owing to differences in passenger mass. If car A approaches at 4.50 ms and car B is moving at 3.70 ms, calculate (a) their velocities after the collision, and (b) the change in momentum of each.

The image in the question is missing and i have attached it here.

A) Final Velocity of Car A = 3.648 m/s

Final velocity of Car B = 4.448 m/s

B) Change in momentum of Car A = 370.62Kg.m/s

Change in momentum of Car B = 370.26 Kg.m/s

Explanation:

A) Mass of car A is 435 kg.

Mass of car B is 495 kg.

Initial speed of car A is 4.5 m/sec.

Initial speed of car B is 3.7 m/sec.

From law of conservation of energy,

M1U1 + M2U1 =M1V1 + M2V2

So in this case;

(435 x 4.5) + (495 x 3.7) = 435V1 + 495V2

And so;

3789 = 435V1 + 495V2 ------- eq 1

We also know that the relative velocity of one particle with respect to the other is reversed by the collision. Thus;

U1 - U2 = - (V1 - V2)

And so, U1 - U2 = V2 - V1

Thus, V2 - V1 = 4.5 - 3.7 = 0.8 m/s

V2 = V1 + 0.8

Putting V1 + 0.8 for V2 in eq 1 to obtain;

3789 = 435V1 + 495(V1 + 0.8)

3789 = 435V1 + 495V1 + 396

3789 - 396 = 930 V1

V1 = 3393/930 = 3.648 m/s

Since V2 = V1 + 0.8

Thus; V2 = 3.648 + 0.8 = 4.448 m/s

B) Momentum of Car A before collision = M1U1 = 435 x 4.5 = 1957.5 Kg.m/s

Momentum of car A after collision is given as;

= M1V1 = 435 x 3.648 = 1586.88 Kg.m/s

we know that change in momentum is given by;

Change in momentum = M1U1 - M1V1

So change in momentum of Car A =

1957.5 - 1586.88 = 370.62Kg.m/s

Similarly;

Momentum of Car B before collision = M2U2 = 495 x 3.7 = 1831.5 Kg.m/s

Momentum of car B after collision is given as;

= M2V2 = 495 x 4.448 = 2201. 76 Kg.m/s

we know that change in momentum is given by;

Change in momentum = M2V2 - M2U2

So change in momentum of Car B=

2201.76 - 1831.5 = 370.26 Kg.m/s