First slide

Second law

Question

A cricket ball of mass 150 g has an initial velocity $u = (3 \hat{i} + 4 \hat{j}) {ms}^{- 1}$ and a final velocity $v = - (3 \hat{i} + 4 \hat{j}) {ms}^{- 1}$ , after being hit. The change in momentum (final momentum - initial momentum) is (in kg-ms^-1)

Moderate

A

zero
B

$- (0.45 \hat{i} + 0.6 \hat{j})$
C

$- (0.9 \hat{i} + 1.2 \hat{j})$
D

$- 5 (\hat{i} + \hat{j}) \hat{i}$

Solution

Given, $u = (3 \hat{i} + 4 \hat{j}) {ms}^{- 1} and v = - (3 \hat{i} + 4 \hat{j}) {ms}^{- 1}$
Mass of the ball = 150 g = 0.15 kg
$∴ ∆ p$ = Change in momentum
= Final momentum - Initial momentum = mv - mu
$= m (v - u) = (0.15) [- (3 \hat{i} + 4 \hat{j}) - (3 \hat{i} + 4 \hat{j})]$
$= (0.15) [- 6 \hat{i} - 8 \hat{j})] = - [0.15 \times 6 \hat{i} + 0.15 \times 8 \hat{j})]$
$= - (0.9 \hat{i} + 1.20 \hat{j})$
$Hence, ∆ p = - (0.9 \hat{i} + 1.2 \hat{j})$

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