3 blocks are initially placed as shown . block A has a mass m and initial velocity v to the right. block B with mass m and block C with mass 4m are initially at rest. (neglecting friction) all collisions are elastic . find the final velocity of blok A.
1) 0.6 v to left
2) 0.4 v to the left
3) v to the left
4) 0.4 v to the right
plz.. give the explanation of the solution
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The velocity of block A is 0.6v towards left.
Explanation:
- When A collides with B, the velocity of A becomes 0, and that of B becomes vv.
- When B moving with velocity vv towards C collides with C.
- Since mass of C is greater, B will bounce back and start moving towards left with velocity v1
- C will start moving with velocity v2 towards left.
- Using law of conservation of momentum:
mv = − mv1 + 4 mv^ 2
V2 = V + V1 / 4
- Now using conservation of energy:
1 / 2 mv^2 = 1/2 mv1^2 + 1 / 2 4mv2^2
- Replacing v2 in the equation we get:
5v1^2 + 2vv1 − 3v2^2 = 0
- Solving the quadratic equation we get:
v1 = − 0.6 v i.e 0.6 v towards left.
Hence the velocity of block A is 0.6 v towards left.
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