** An object of mass m1=5.9 kg moving at 4.7 m/s strikes a stationary second obj

Question

** An object of mass m1=5.9 kg moving at 4.7 m/s strikes a stationary second object of unknown mass. After an elastic collision, the first object is observed moving at 3.76 m/s at an angle of -33° with respect to the original line of motion.

What is the energy of the second object? 23.46 J

What is the magnitude of the second object's momentum after the collision?

At what angle did the second ball move relative to the direction of the first ball's initial motion?

What is the mass of the second ball?

Explanation / Answer

using momentum conservation in both axis

m1v1=m1v1'cos33+m2v2'cosB (In X direction)

0=-m1v1'sin33+m2v2'sinB (In Y direction)

Using conservation of energy we have that:

v1'=(m1-m2)v1/(m1+m2)=(5.9-m2)*4.7/(5.9+m2)=3.76

solving for m2=0.655 kg

Using conservation of energy we have that:

v2'=2m1*v1/(m1+m2)=2*5.9*4.7/(5.9+0.655)=8.46 m/s

The magnitude of the momentum is m2v2'=0.655*8.46=5.5413 kg m/s

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