A 3.70 g bullet moving at 250 m/s enters and stops in an initially stationary 2.

ID: 1417103 • Letter: A

Question

A 3.70 g bullet moving at 250 m/s enters and stops in an initially stationary 2.00 kg wooden block on a horizontal frictionless surface.

1.) What's the speed of the bullet/ block combination?

Express your answer to two significant figures and include the appropriate units.

2.) What fraction of the bullet's kinetic energy was lost in this perfectly inelastic collision?

3.) How much work was done in stopping the bullet?

Express your answer to two significant figures and include the appropriate units

4.) If the bullet penetrated 5.00 cm into the wood, what was the average stopping force?

Express your answer to two significant figures and include the appropriate units.

m1 = mass of bullet = 3.70 g = 0.0037 kg

V1i = initial velocity of the bullet = 250 m/s

m2 = mass of block = 2 kg

V2i = initial velocity of the bullet = 0 m/s

V = Velocity of the combination

Using conservation of momentum

m1 V1i + m2 V2i = (m1 + m2) V

0.0037 x 250 + 2 x 0 = (0.0037 + 2) V

V = 0.46 m/s

fraction lost = Final KE / initial KE = (0.5) (m1 + m2) V2 /((0.5) m1 V1i2) = (0.0037 + 2) (0.46)2 /((0.0037) (250)2) = 0.0018

Work done = Change in KE of bullet = - (0.5) m1 V21i = - (0.5) (0.0037) (250)2 = - 115.625 J

F d = 115.625

F (0.05) = 115.625

F = 2312.5 N = 2.3 x 103 N

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