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Conservation of Momentum

A cricket bal...

Question

A cricket ball of mass 100g has an initial velocity $u = (3 i + 4 j) m / s$ and a final velocity $v = - (3 i + 4 j) m / s$ after being hit.
Find the change in momentum.

$0$

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$- (0.45 i + 0.6 j)$

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$- (0.9 i + 1.2 j)$

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$- 5 (i + j)$

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Solution

The correct option is C
$- (0.9 i + 1.2 j)$

Step 1: Given data-
Mass of ball $m$ = $150 g = 0.15 k g .$
Initial velocity $u equals left parenthesis 3 space i space plus 4 space j right parenthesis space m divided by s space$
Final velocity $v equals negative open parentheses 3 space i plus space 4 space j close parentheses space m divided by s$ .
Let $P d e n o t e s m o m e n t u m .$
Step 2: Finding change in momentum $∆ P$ .
$∆ P = m v - m u = m (v - u) = 0.15 (- (3 i + 4 j) - (3 i + 4 j)) = - \{0.9 i + 1.2 j\} k g m / s$
Hence, option C is correct.

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Similar questions

150 g

u = 3^i + 4 j m / s

v = - (3^i + 4 j) m / s

A body of mass 150 g has an initial velocity (3i+4j)m/s and final velocity -(3i +4j) m/s after being hit. Calculate the change in momentum.

3^i + 4^j

(4^i - 3^j) m / s^{2}

(3 ˆ i + 4 ˆ j) m / s

(0.4 ˆ i + 0.3 ˆ j) m / s^{2}

.

10 s .

m = 0

1 k g

3^i m / s

B

5^j m / s

\to v = 4 (^i +^j)

B

\frac{x}{10} J

x

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