Impulse # 2

A system of two blocks A and B are connected by an inextensible massless string as shown in above figure. The pulley is massless and frictionless . Initially , the system is at rest .

A bullet of mass m m moving with a velocity u u as shown above hits block B and gets embedded into it . The impulse imparted by tension force to the block A is ?

4 m u 5 \dfrac{4mu}{5} 3 m u 5 \dfrac{3mu}{5} 2 m u 5 \dfrac{2mu}{5} 5 m u 4 \dfrac{5mu}{4}

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1 solution

Rishu Jaar
Nov 12, 2017

By conservation of Linear momentum along the string ,

m u = ( m + m + 3 m ) v v = u 5 ( v is the common instantaneous velocity just after the collison ) mu = (m+m+3m)v ~ \implies v = \dfrac{u}{5} ~~~~~~~~\small \color{#20A900}{( \text{v is the common instantaneous velocity just after the collison} )}

Therefore the impulse on block A = I I = 3 m ( v 0 ) = 3 m u 5 ~3m(v - 0) = \color{#3D99F6}{ \boxed{\dfrac{3mu}{5}}}

Yeah! Same way!! :)

Md Zuhair - 3 years, 7 months ago

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Nice :) ; Hey could you send me an invite for the slack group.

Rishu Jaar - 3 years, 7 months ago

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Which group? I am not in that JEE Group

Md Zuhair - 3 years, 7 months ago

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@Md Zuhair oh , i thought Harsh , Swapnil , and you were active with IITians at the slack group. Still, its ok . But if possible , please inform me , i could sure use some help.

Rishu Jaar - 3 years, 7 months ago

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