Question

A steel ball A, of mass 20.0 kg moving with a speed of 1 2.0ms− collides with another ball B of mass 10.0 kg which is initially at rest. After collision, A moves with a speed of 1 1.0ms− at an angle of 30º with its original direction of motion. Determine the final velocity of B.

Answer 1

Answer:

From the conservation of momentum:

Mv_i=Mv_f \cos{30}+mu\cos{a}Mv

i

=Mv

f

cos30+mucosa

0=Mv_f \sin{30}-mu\sin{a}0=Mv

f

sin30−musina

Thus:

u=v_f \frac{M \sin{30}}{m\sin{a}}u=v

f

msina

Msin30

Mv_i=Mv_f \cos{30}+mv_f \frac{M \sin{30}}{m\sin{a}}\cos{a}Mv

i

=Mv

f

cos30+mv

f

msina

Msin30

cosa

20(2)=20(1) \cos{30}+20(1)\sin{30}\cot{a}20(2)=20(1)cos30+20(1)sin30cota

The direction is 23.8 degrees from original direction of motion in another side from steel ball A .

The speed is

u=1 \frac{20 \sin{30}}{10\sin{23.8}} =2.48 m/su=1

10sin23.8

20sin30

=2.48m/s