A cricket ball of mass `150g` has an initial velocity `(3 hati + 4hatj)ms^(-1)` and a final velocity `upsilon = -(3hati + 4hatj)ms^(-1)` after beigh hit The change in momentum (final momentum initial momentum) is (in kg `ms^(-1)`)

Options

A. zero
B. `-(0.45 hat(i)+0.6 hat(j))`
C. `-(0.9 hat(i)+1.2 hat(j))`
D. `-5(hat(i)+hat(j))hat(i)`

Answer

C

Explanation

Given, `u=(3hat(i)+4 hat(j)) ms^(-1)` and `v=-(3 hat(i)+4 hat(j)) ms^(-1)`
Mas of the ball `=150 g=0.15 kg`
`Delta p=` Change in momentum
=Final momentum -Intial momentum
`=mv-m u`
`=m(v-u)=(0.15) [-(3 hat(i)+4 hat(j))-(3hat(i)+4hat(j))]`
`=(0.15) [-6 hat(i)-8hat(j)")"]=-[0.15xx6 hat(i)+0.15 xx8 hat(j)")"]`
`=-[0.9 hat(i)+1.20 hat(j)]`
Hence, `Delta p=-[0.9 hat(i)+1.2 hat(j)]`