a bullet of mass 1 = 0.0065kg that travels horizontally to the right at v1= 710

Question

a bullet of mass 1 = 0.0065kg that travels horizontally to the right at v1= 710 m/s is fired into and passes through mass 2 = 0.2 kg wood block which is initially moving to left on a horizontal frictionless surface with speed v2= 250 m/s.
A. Find velocity of block after collision
B. if collision time is 0.0370 seconds, find average force exerted on bullet by the block
C. Find max height above ground to which the block rises after collision a bullet of mass 1 = 0.0065kg that travels horizontally to the right at v1= 710 m/s is fired into and passes through mass 2 = 0.2 kg wood block which is initially moving to left on a horizontal frictionless surface with speed v2= 250 m/s.
A. Find velocity of block after collision
B. if collision time is 0.0370 seconds, find average force exerted on bullet by the block
C. Find max height above ground to which the block rises after collision a bullet of mass 1 = 0.0065kg that travels horizontally to the right at v1= 710 m/s is fired into and passes through mass 2 = 0.2 kg wood block which is initially moving to left on a horizontal frictionless surface with speed v2= 250 m/s.
A. Find velocity of block after collision
B. if collision time is 0.0370 seconds, find average force exerted on bullet by the block
C. Find max height above ground to which the block rises after collision

Explanation / Answer

v1 = (m1-m2)*u1/(m1+m2) - 2*m2*u2/(m1+m2)

v1 = ((0.0065-0.2)*710)/(0.0065+0.2) - (2*0.2*250)/(0.0065+0.2) = -1149.56 m/s

v2 = 2*m1*u1/(m1+m2) - (m2-m1)*u2/(m1+m2)

v2 = (2*0.0065*710)/(0.0065+0.2) - ((0.2-0.0065)*250)/(0.2+0.0065)

v2 = -189.56 m/s

A) velocity of block after collision is v2 = 189.56 to the left

B) F*dt = 0.0065(710+1149.56) = 1859.56

F = 1859.56/0.037 = 50258.4 N

C) maximum height Hmax = v2^2/(2*g) = 189.56*189.56/(2*9.81) = 1831.44 m

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