Question

The 4kg head of a sledge hammer is moving at 6m/s when it strikes a chisel, driving it into a log. The duration of the impact (or the time for the sledge hammer to stop after contact) is 0.0020s. Find the time average of the impact force

Answer 1

Answer:

The average impact force is 12000 newtons.

Explanation:

By Impact Theorem we know that impact done by the sledge hammer on the chisel is equal to the change in the linear momentum of the former. The mathematical model that represents the situation is now described:

$\bar F \cdot \Delta t = m \cdot (v_{2}-v_{1})$ (1)

Where:

$\bar F$ - Average impact force, in newtons.

$\Delta t$ - Duration of the impact, in seconds.

$m$ - Mass of the sledge hammer, in kilograms.

$v_{1}$, $v_{2}$ - Initial and final velocity, in meters per second.

If we know that $\Delta t = 0.0020\,s$, $m = 4\,kg$, $v_{1} = -6\,\frac{m}{s}$ and $v_{2} = 0\,\frac{m}{s}$, then we estimate the average impact force is:

$\bar F = \frac{m\cdot (v_{2}-v_{1})}{\Delta t}$

$\bar F = 12000\,N$

The average impact force is 12000 newtons.