An object of mass m 1 =7.1 kg moving at 5.3 m/s strikes a stationary second obje

Question

An object of mass m1=7.1 kg moving at 5.3 m/s strikes a stationary second object of unknown mass. After an elastic collision, the first object is observed moving at 3.71 m/s at an angle of -41° with respect to the original line of motion.
1. What is the energy of the second object? 50.86 J

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

3. At what angle did the second ball move relative to the direction of the first ball's initial motion? (Give answer in degrees)

4. What is the mass of the second ball?

Explanation / Answer

2) Magnitude of second object's momentum

m1u1 = m1v1cos1 + m2v2cos2

[ (7.1 kg) * (5.3 m/s) ] = [ (7.1 kg) * (3.71 m/s) * (cos41) ] + m2v2cos2

[ 37.63 kg-m/s ] = [ (26.341 kg-m/s) * (0.754) ] + m2v2cos2

[ 37.63 kg-m/s ] - [ 19.86 kg-m/s ] = m2v2cos2

m2v2cos2 = 17.77 kg-m/s

Conserve vertical momentum:
0 = 7.1kg * 3.71m/s * sin(-41) + m2 * v2 * sin2
m2v2sin2 = 17.28 kg·m/s

Divide vertical momentum by horizontal momentum:
m2v2sin2 / m2v2cos2 = tan2 = 17.77kg·m/s / 17.28 kg·m/s = 1.028
2 = arctan1.028 = 45.79 º ( angle of second ball)

m2v2cos45.79 = 17.77 kg·m/s
MV = 25.484 kg·m/s
Divide into KE:
½MV² / MV = V/2 = 50.86J / 25.484 kg·m/s = 1.995 m/s
V = 3.991 m/s

50.86J = ½*m2*(3.991m/s)²
m2 = 6.389 kg mass of second ball

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