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.
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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
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