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. `-(0.45hati + 0.6hatj)`
C. ` -(0.9hati + 1.2hatj)`
D. ` -5(hati + hatj)`

Answer

c

Explanation

Here, `m = 150g = 0.15kg , vec(u) = (3hati + 4hatj)m//s, vec(upsilon) = - (3hati + 4hatj)m//s`
Change in momentum `vec (p) = m vec(upsilon) - m vec(upsilon) = 0.15 [-(3hati + 4hatj) -(3hati + 4hatj)] = - 0.30 [3hati + 4hatj)]`
`vec(p) = - (0.9hati + 1.2hatj) kgm//s` .