A 10,000 kg coal car moving at 3.60 m/s (i) collides with a 16000 kg coal car mo

Question

A 10,000 kg coal car moving at 3.60 m/s (i) collides with a 16000 kg coal car moving at 1.00 m/s (-i).

(a) If the cars couple upon collision, what is the change in momentum of the 10,000 kg coal car? and its
direction.
The answers are (a) 36,000 Ns (b) 3,690 Ns (c) 16,000 Ns (d) 12,300 Ns (e) 28,300 Ns (f) 7,690 Ns

(b) What is the change in momentum of the 16,000 kg coal car? and its direction.
The answers are (a) 36,000 Ns (b) 3,690 Ns (c) 16,000 Ns (d) 12,300 Ns (e) 28,300 Ns (f) 7,690 Ns


PLEASE SHOW WORK, Thanks!

Explanation / Answer

To answer this you need to

1. find the final velocity of both cars, using cons of momentum

2. Find the final momentum of each car

3. Find the change in momentum of each car

First:    initial momentum = final momentum

            10000 * 3.6 + 16000 * -1 = (10000 + 16000) * final velocity

              20000 = 26000 * final velocity

    final velocity = 0.7692 m/s (in + direction)

Second: final momentum of Car 1 is   10000*0.7692 = 7692

              final momentum of Car 1 is    16000 * 0.7692 = 12308

Finally, change in momentum for car 1 is

                final - initial = 7692 - 36000 = -28300 Ns (so 28300 Ns in the - direction)

           for car 2

             final - initial = 12308 - -16000 =   28300 Ns    (in the + direction)

(Note: they must have equal and opposite changes in momentum... conservation of momentum)

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