An object of mass m 1 =7.1 kg moving at 4.7 m/s strikes a stationary second obje

Question

An object of mass m1=7.1 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 2.82 m/s at an angle of -43° with respect to the original line of motion.
What is the energy of the second object?

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? (Give answer in degrees)

What is the mass of the second ball?

Explanation / Answer

0.5m1vo^2=0.5m1v1^2+0.5m2v2^2

0.5m2v2^2=0.5m1(vo^2-v1^2)=50.19 J which is the energy of the second object

Linear momentum in x:

m1vo=m1v1cos43°+m2v2

7.1*4.7=7.1*2.82*cos43 + m2v2

m2v2cosø=18.73

Linear momentum in y:

0=m1v1sin43°+m2v2sinø

m2v2sinø=13.65

P2=sqrt(18.73^2+13.65^2)=23.18

The angle is m2v2sinø/m2v2cosø =sinø/cosø=tgø= 13.65/18.73

angle tg^-1(0.7288)=36.08°

The mass is m=P2^2/2K2 = 23.18^2/2*50.19=5.35 kg

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