A body of mass 2 kg has an initial velocity vi- ms-1- After the collision with another body its velocity becomes vf-5 -6 -k-ms-1- If the impact time is 0-02 s- the average force of impact on the body in newton is

Change in momentum = m(v_f v_i)=2 kg(5 î+ 6 ĵ k̂) ms^ 1 2 kg(î+ ĵ) ms^ 1=2(4 î+ 5 ĵ k̂) Average force = change in momentum/time of impact =2 (4 î+5 ĵ k̂)/0 ...

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Byju's Answer

Standard XIII

Physics

Impulsive Forces in Collision

A body of mas...

Question

A body of mass 2 kg has an initial velocity $v_{i} = (^i +^j) m s^{- 1}$ . After the collision with another body its velocity becomes $v_{f} = (5^i + 6^j -^k) m s^{- 1}$ . If the impact time is 0.02 s, the average force of impact on the body (in newton) is

50 (4^i + 5^j +^k)

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50 (4^i - 5^j -^k)

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100 (4^i + 5^j -^k)

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100 (4^i + 5^j -^k)

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Solution

The correct option is D $100 (4^i + 5^j -^k)$
Change in momentum = $m (v_{f} - v_{i}) = 2 k g (5^i + 6^j -^k) m s^{- 1} - 2 k g (^i +^j) m s^{- 1} = 2 (4^i + 5^j -^k)$
Average force = $\frac{c h a n g e i n m o m e n t u m}{t i m e o f i m p a c t}$
= $\frac{2 (4^i + 5^j -^k)}{0.02 s} = 100 (4^i + 5^j -^k)$
Hence the correct choice is (d).

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A body of mass 2 kg has an initial velocity vi=(^i+^j)ms−1. After the collision with another body its velocity becomes vf=(5^i+6^j−^k)ms−1. If the impact time is 0.02 s, the average force of impact on the body (in newton) is

The correct option is D 100(4^i+5^j−^k)Change in momentum = m(vf−vi)=2 kg(5^i+6^j−^k) ms−1 −2 kg(^i+^j) ms−1=2(4^i+5^j−^k) Average force = change in momentumtime of impact =2(4^i+5^j−^k)0.02 s=100(4^i+5^j−^k) Hence the correct choice is (d).

A body of mass 2 kg has an initial velocity $v_{i} = (^i +^j) m s^{- 1}$ . After the collision with another body its velocity becomes $v_{f} = (5^i + 6^j -^k) m s^{- 1}$ . If the impact time is 0.02 s, the average force of impact on the body (in newton) is