# Momentum physicsThe mass of the hammer is 0.75 kg and its velocity, just before it hits the nail, is 15.0 m/s downward. After hitting the nail, the hammer remains in contact with it for 0.1 s....

The mass of the hammer is 0.75 kg and its velocity, just before it hits the nail, is 15.0 m/s downward. After hitting the nail, the hammer remains in contact with it for 0.1 s. After this time both the hammer and the nail have stopped moving.

What is the momentum of the hammer just before it hits the nail?

Show how you work out your answer and give the units and direction.

Momentum = ...................................................................

(3)

(iii) What is the change in momentum of the hammer during the time it is in contact with the nail?

(1)

(iv) Write down an equation which connects change in momentum, force and time.

(1)

(v) Calculate the force applied by the hammer to the nail.

Show how you work out your answer and give the unit.

Force = ............................................................................

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Momentum just before hitting = mass* velocity =0.75kg*15=11.25kgm/s in the direction of velocity (down)

Change in momentum= mass *velocity after hit - mass * velocity before hit= 0.75(0m/s-15m/s) = - 11.25kgm/s in the direction of velocity or 11.25kgm/s in the opposite direction of veocity of the hammer.

Change in momentum/time = rate of change in momentum = Force. Therefore,

Force, F = mv-mu/t = m(v-u)/t, where m is the mass of the hammer, v and u are the final velocity after and before hit. The time of contact after hit is t =0.1s.Therefore,

F = 0.75kg(0m/s-15m/s)/(0.1s) = -112.5 kgm/s^2 = -112.5 N is the force in the direction of velocicty of the hammer that stopped the hammer. Or 112.5 N in a direction opposite to the velocity of the hammer before the hit.

Since the applied force + the force F that stopped the hammer = 0, Applied force = -F = -(-112.5) = 112.5N

Applied force on the nail = 112.25 N in the direction of the velocity of the hammmer before hit.