Question

A hammer of mass $1$kg moving with a speed of $6$m/s strikes a wall and comes to rest in $0.1$sec. Calculate:
1. An impulse of the force.
2. An average retarding force that stops the hammer.
3. Average retardation of the hammer.

Answer 1

Hint: Impulse is the force applied for small intervals of time. It is equal to change in momentum.
1. Impulse $ = F \times t$
$ = $ Force $ \times $ time interval
2. Impulse $ = $ Change in momentum
$ = $ final momentum – initial momentum
3. Force $ = $ mass $ \times $ acceleration

Complete step by step answer:
Mass of the hammer, $m = 1Kg$
Initial speed of hammer $ = $ speed with which it strikes the wall, $u = 6m/s$
Final speed, v$ = 0$m/s
Time taken, $t = 0.1\sec $.
1. Impulse of force $ = $ change in momentum $ = $ final momentum – initial momentum…… (i)
Final momentum$ = $mv$ = 1\left( 0 \right) = 0$kg m/s
Initial momentum$ = mu = 1\left( 6 \right) = 6$kg m/s
So, putting these in equation (i), we have impulse of force $ = 0 - 6$
$ = - 6$kg m/s
$ = - $6 N
2. as impulse $ = $ force $ \times $ time interval
So, the retarding force that stops a Hammer $ = \dfrac{{impulse}}{{time{\text{ interval}}}}$
$ = - \dfrac{6}{{0.1}} = - 60N$
So, retarding force $ = 60N$
3. By Newton’s law,
Force $ = $ mass $ \times $ acceleration
So, acceleration $ = \dfrac{{Force}}{{Mass}} = \dfrac{{ - 60N}}{1}$
$ = - 60m{s^{ - 2}}$
So, retardation of hammer $ = 60m{s^{ - 2}}$
Additional information: We know that impulse, $J = F \times \Delta t$and by newton’s second law of motion, force is given by
$F = \dfrac{{\Delta p}}{{\Delta t}}$
So, impulse, \[J = \dfrac{{\Delta p}}{{\Delta t}} \times \Delta t\]
$J = \Delta p$
Or impulse $ = $ change in momentum.

Units of impulse:
As impulse is a change in momentum, SI units of impulse are the same as that of momentum.

Note:
Although acceleration $ = - 60m{s^{ - 2}}$
But retardation is $ = 60m{s^{ - 2}}$
This is because the retardation itself means negative acceleration. So, negative sign is not used while writing retardation.