Question

2 fig. 2.1 shows a hammer being used to drive a nail into a piece of wood. the mass of the hammer head is 0.15 kg. the speed of the hammer head when it hits the nail is 8.0 m/s. the time for which the hammer head is in contact with the nail is 0.0015 s. the hammer head stops after hitting the nail. (a) calculate the change in momentum of the hammer head. change in momentum = ………………………………………… 2 (b) state the impulse given to the nail. impulse = ………………………………………… 1 (c) calculate the average force between the hammer and the nail. average force = ………………………………………… 2 total: 5

Explanation:

Step1: Calculate the change in momentum for part (a)

The formula for momentum is \(p = mv\). The initial velocity \(v_i=8.0\ m/s\) and the final velocity \(v_f = 0\ m/s\). The mass \(m = 0.15\ kg\).
The change in momentum \(\Delta p=m(v_f - v_i)\)
Substitute the values: \(\Delta p=0.15\times(0 - 8.0)\)
\(\Delta p=- 1.2\ kg\cdot m/s\). The magnitude of the change in momentum is \(1.2\ kg\cdot m/s\)

Step2: State the impulse for part (b)

According to the impulse - momentum theorem, \(J=\Delta p\). Since the force exerted by the hammer on the nail and the force exerted by the nail on the hammer are an action - reaction pair, the impulse given to the nail is equal in magnitude (opposite in direction) to the change in momentum of the hammer. So \(J = 1.2\ N\cdot s\)

Step3: Calculate the average force for part (c)

The formula for impulse is \(J = F_{avg}\times\Delta t\). We know \(J = 1.2\ N\cdot s\) and \(\Delta t=0.0015\ s\)
Rearrange the formula for \(F_{avg}\): \(F_{avg}=\frac{J}{\Delta t}\)
Substitute the values: \(F_{avg}=\frac{1.2}{0.0015}\)
\(F_{avg}=800\ N\)

Answer:

(a) \(1.2\ kg\cdot m/s\)
(b) \(1.2\ N\cdot s\)
(c) \(800\ N\)