Newton's

Classical Mechanics Level 3

A ball of mass 0.2 kg is normally thrown against the wall with a speed of 15m/s and rebounds from there with a velocity 20m/s. The impulse of the force exerted by the ball on the wall in 'N s' is-


The answer is 7.

Harsh Shah by Harsh Shah , 20, India

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2 solutions

From Newton's Second Law, we get that the rate of change of momentum is equal to the force on the object. That is :

F = \frac{Δp}{Δt}

Hence, F × Δ t = Δ p F\times Δt = Δp

→ J = Δ p = ( 0.2 ) ( 20 ) − ( − 15 ) ( 0.2 ) = 4 + 3 = 7 J = Δp = (0.2)(20) - (-15)(0.2) = 4 + 3 = \boxed{7}

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YUMMY TASTY QUESTION......

ashutosh mahapatra - 6 years, 9 months ago

Did the same

Parth Lohomi - 6 years, 9 months ago

dei this one is a vey easy one da

Guru Prasaadh - 6 years, 3 months ago
Anchal Rajawat
Aug 1, 2014

Change in momentum=mv-mu here,v=-20m/s, u=15m/s impulse=0.2 (-20)-0.2 15= -7 this negative sign indicates direction.... so, answer is 7

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