An iron ball of mass `m=50 g` falls from a height of `(h_1)=5m` and rises upto `h_(2)=3.2m` after colliding with the horizontal surface. If the time of contact with the surface is `Delta t=0.02s`, find the average contact force exerted on the ball by the horizontal surface.

Options

A. `-45 hatj N`
B. `45 hatj N`
C. `45 hatiN`
D. `-45 hati N`

Answer

B

Explanation

The change in linear momentum during collision is
`Delta vec(p)=mvec(v)_(2)-mvec(v)_(1)`
`=m(vec(v_2)-vec(v_1))`
`=m[v_(2)hat(j)-(-v_(1)hat(j))]=m[(v_(1)+v_(2))]hat(j)`
The average force during the collision is
`vec(F)=(Delta vec(p))/(Delta t)=(m(v_(1)+v_(2))hat(j))/(Delta t)`

We can calculate `v_(1) and v_(2)` by using kinematics .
Using `v^(2)=u^(2)+2as`
Putting a=g and s=h, we get
`v_(1)=sqrt(2gh_(1)), (v_2)=sqrt(2gh_(2))`
or, `vec(F)=(m(sqrt(2gh_(1))+sqrt(2gh_(1)))hat(j))/(Delta t)`
`=((50)/(1000)) ((sqrt(2 xx 10 xx 5)+sqrt(2 xx 10 xx3.2)))/(0.02) hat(j)=45hat(j)N`.
During collision, the horizontal surface pushes the ball up with an average force of 45N.