A cricket ball of mass 150 g has an intial velocity u = `(3 hati + 4 hatj)ms^(-1)` and a final velocity `v = -(3hati + 4 hatj) ms^(-1)` , after being hit. The change in momentum (final momentum - initial momentum ) is (in `Kg ms^(1)`)

Options

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

Answer

C

Explanation

Given, `u = (3hat(i) + 4hat(j))m//s`
and `v = - (3 hat(i) + 4 hat(j))m//s`
Mass of the ball = 150g = 0.15g
`Delta p` = Change in momentum = Final momentum - Initial momentum
= mv - mu
` = m(v-u) = (0.15)[(-3 hat(i) - 4hat(j)) - (3hat(i) + 4hat(j))]`
` = 0.15[ - 6 hat(i) - 8hat(j)]`
` = - [0.15 xx 6hat(i) + 0.15 xx 8 hat(j)]`
`[0.9hat(i) + 1.20 hat(j)]`
Hence ,` Delta p = - [0.9 hat(i) + 1.2hat(j)]`