A 27.6 g ball of clay traveling east at 2.22 m/s collideswith a 27.62 g ball of

Question

A 27.6 g ball of clay traveling east at 2.22 m/s collideswith a 27.62 g ball of clay traveling north at 3.76 m/s.
What is the speed of the resulting 55.22 g ball ofclay? What is the direction (angle with respect tox-axis) of the resulting 55.22 g ball of clay? A 27.6 g ball of clay traveling east at 2.22 m/s collideswith a 27.62 g ball of clay traveling north at 3.76 m/s.
What is the speed of the resulting 55.22 g ball ofclay? What is the direction (angle with respect tox-axis) of the resulting 55.22 g ball of clay?

Explanation / Answer

Let both of them after collision go at angle NofE, Applying the law of conservation of momentum along X-axisas m1u1+m2u2=m1v1+m2v2 0.0276*2.22+0.02762*0=55.22*vx vx=0.00110096m/s Along Y-axis we have as, 0.0276*0+0.02762*3.76=55.22*vy vy=0.00188068m/s The resulting speed then as, v=vx2+vy2 v=0.001100962+0.001880682 v=0.002179326934458m/s The resultant direction as, v=0.002179326934458m/s The resultant direction as, tan=vy/vx tan=0.00188068/0.00110096 =59.6550687278354468 NE Hence we get by it.

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