Please help me !

Question

A 15 kg hammer strikes a nail at a velocity of 8.5 m/s and comes to rest in 6.0 ms.

a) What is the impulse given to the nail?  

b) What is the average force acting on the nail?  

Express answers using two significant figures.

  

Answer 1

Let

m = 15 kg be the mass of the hammer,

v₀ = 8.5 m/s be its initial velocity,

v₁ = 0 m/s be its final velocity,

Δt = 6 ms = 0.006 s be the time it takes to change the velocity from v₀ to v₁.

a)

Impulse is defined as

FΔt

where F is a force and Δt the the time period the force is acting.

We do not know the force F. (That will be computed in question b).)

But we know from Newton's Second Law that

FΔt = Δp (1)

where Δp is the change in momentum during that time Δt.

This is true under the assumption that the force remains constant during the time Δt,

which we are going to assume.

We are also going to assume that the mass of the nail is negligible in comparison to

the mass of the hammer.

We can calculate the change in momentum.

Δp = mv₁ - mv₀ = 15×0 - 15×8.5 = -127.5 kgm/s

That is the change in momentum of the hammer, and therefore also the impulse given to the hammer by the nail.

It is negative because the force of the nail acting on the hammer is in direction opposite to the velocity,

which then causes deceleration.

By Newton's third law, the force by which the hammer acts on the nail is equal in magnitude, but opposite in direction.

Therefore the impulse given to the nail by the hammer is 127.5 kgm/s.

b)

Having calculated Δp, we can calculate the average force F from (1)

F = Δp/Δt = 127.5/0.006 = 21250 N