Question

can anyone answer this question
that
A hammer of mass 500g, moving at 50m/s, strikes a nail. The nail stops the hammer In a very short time of 0.01 sec. What is the force of the nail on the hammer. help me friends........

Answer 1

When we all know the famous F=ma equation, but there is another form of Newton's second law, i.e F=dP/dt
i.e force is equal to rate of change of linear momentum.
Here initial momentum of hammer
= mv = 0.5 * 50 =25 Ns
final momentum of hammer = 0 Ns
change in momentum is 0-50= -50 Ns
this change is brought in a very small time 0.01 sec, which can be taken as small change in time t(dt)
F on hammer = -50/0.01 = -5000 N
here minus sign just shows that the force ON hammer is in the opposite direction of it's motion.

Now note this force was provided by the nail and is the only force which changed it's momentum

so Force on hammer due to nail= -5000N
now according to Newton's third law
Forcw on Nail due to Hammer
=- Force on Hammer due to Nail
=-(-5000)
=5000N

Positive value means force is in the direction of motion of hammer.

Hope it helped, feel free to comment.