Question

03
g
Q. 3 A cricket ball of mass 150 g has an initial velocity u = (3î +4ĵ)ms-1 and a
final velocity v=- (3î + 4ſ) ms
ma-1
after being hit. The change in
momentum (final momentum-initial momentum) is (in kgms-2)
(b) – (0.45 +0.69)
(c) – (0.99 +1.2h
(d) – 5( + h) îs sit abeum (b)
no
(a) zero
bin

samjhao guyzzz​

Answer 1

$\green{\rm \implies{ \underline{given}}}$

cricket ball of mass = 150 g

$\rm \to \: initial \: velocity \: = u \: = (3\hat{i} + 4 \hat{j})ms {}^{ - 1}$

$\rm \to \: final \: velocity \: = \: v \: = - (3 \hat{i} + 4 \hat{j})ms {}^{ - 1}$

$\blue{ \underline{ \rm \to \: to \: find \: the \: change \: in \: momentum}}$

According to the question,

mass = 150g = 0.15 kg.

Let,

$\rm \to \: initial \: momentum \: = \vec{p}_{i} = mu$

$\rm \to \: (0.15)(3 \hat{i} \: + 4 \hat{j})ms {}^{ - 1}$

$\rm \to \: (0.45 \hat{i} \: + 0.6 \hat{j}) \: kgms {}^{ - 1}$

$\rm \to \: final \: momentum \: = \vec{p}_{f} \: = \: mv$

$\rm \to \: (0.15)( - 3 \hat{i} - 4 \hat{j})ms {}^{ - 1}$

$\rm \to \: ( - 0.45 \hat{i} - 0.6 \hat{j})kgms {}^{ - 1}$

$\rm \implies{ \underline{change \: in \: momentum \: = \Delta p \: = p_{f} \: - \: p_{i} }}$

$\rm \to \: ( - 0.45 \hat{i} - 0.6 \hat{j}) - (0.45 \hat{i} + 0.6 \hat{j})$

$\rm \to - (0.9 \hat{i} + 1.2 \hat{j})kgms {}^{ - 1}$