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Question

Three blocks are initially placed as shown in the figure. block $$A$$ has mass $$m$$ and initial velocity $$v$$ to the right. Block $$B$$ with mass $$m$$ and block $$C$$ with mass $$4m$$ are both initially at rest. Neglect friction. All collisions are elastic. The final velocity of block $$A$$ is :

133015_ffc0b6a999ce4d1aa5d6d26a4d7a8fe3.jpg


A
0.6v to the left
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B
1.4v to the left
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C
v to the left
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D
0.4v to the right
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Solution

The correct option is C $$0.6v$$ to the left
When A collides with B, the velocity of A becomes 0, and that of B becomes $$v$$.
When B moving with velocity $$v$$ towards C collides with C, since mass of C is greater, B will bounce back and start moving towards left with velocity. say $${v}_{1}$$ and C will start moving with velocity, say $${v}_{2}$$ towards left. 
So, using law of conservation of momentum:
$$mv=-m{v}_{1}+4m{v}_{2}$$
$${v}_{2}=\dfrac{v+{v}_{1}}{4}$$
Now using conservation of energy:
$$\dfrac{1}{2}m{v}^{2}=\dfrac{1}{2}m{{v}_{1}}^{2}+\dfrac{1}{2}4m{{v}_{2}}^{2}$$
Replacing $${v}_{2}$$ in the equation we get:
$$5{{v}_{1}}^{2}+2v{v}_{1}-3{{v}_{2}}^{2}=0$$
Solving the quadratic equation we get:
$${v}_{1}=-0.6v$$ i.e $$0.6v$$ towards left
When B collides with A again, A moves with velocity $${v}_{1}$$, thus final velocity of block A is $$0.6v$$ towards left.

Physics

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