Question

a body of mass 5 kg strike the floor with a speed of 2 m s if it rebounds with a speed of 1 MS and the time of contact with the floor is for 5 seconds then find the force of impact in Newton​

Answer 1

Given : m = 5kg

initial v = 2 m/s

final v = 1 m/s

t =5s

To Find : force of impact i.e force exerted

As we know ,

Force exerted  = rate of change of momentum

F=$\frac{dP}{dt}$

we are given action reaction pair  with which the body strikes and rebounds.

the ball initially strikes in +ve x direction with

initial momentum $P_i$ = 5*2 = 10kgm/s

And rebounds in -ve x direction with

final momentum $P_{f}$ = 5*(-1) = -5kgm/s

$\Delta P\ =\ P_i-P_f$

= 10-(-5) = 10+5= 15kgm/s

NOW,

$\Delta F\ =\frac{\Delta P}{\Delta t}$

=$\frac{15}{5}$

$\Delta F\ =\ 3N$

Answer 2

Given: Mass of the body = 5kg

          Velocity before collision, u = 2m/s

          Velocity after collision, v = -1m/s

To find: The force of impact in Newton

Solution:

Acceleration is the rate of change of velocity.

The acceleration produced by the body = {u - (-v)}/t

                                                                   = 2 - (-1)/5

                                                                   = 3/5 ms⁻²

Force of impact = mass × acceleration

                          = 5×3/5

                          = 3N

Therefore, the force of impact in Newton is 3N.

• The negative in the sign is due to the direction being taken as negative against the gravity, while the one which is towards the gravity is taken a positive.
• Force acting on the body is equal to the product of the mass of the body and the acceleration the force produces. Its SI unit is Newton.