A block of mass 3 M at rest on a frictionless horizontal surface explodes into two pieces. One piece of mass M moves with speed V at an angle of 6 0 ∘ with the original direction of motion (which was

Question

A block of mass 3 M at rest on a frictionless horizontal surface explodes into two pieces. One piece of mass M moves with speed V at an angle of 6 0 ∘ with the original direction of motion (which was zero). What is the speed of the other piece?

TestHub · Accepted Answer

Since the block is at rest on a frictionless surface, the initial momentum of the system is zero. By conservation of momentum, the total momentum after the explosion must also be zero.

Let the mass of the first piece be $m_{1} = M$ and its velocity be $v_{1}$ . Let the mass of the second piece be $m_{2} = 3 M - M = 2 M$ and its velocity be $v_{2}$ . Initial momentum: $P_{i} = 0$ . Final momentum: $P_{f} = m_{1} v_{1} + m_{2} v_{2} = M v_{1} + 2 M v_{2}$ . From momentum conservation, $M v_{1} + 2 M v_{2} = 0$ , which implies $M v_{1} = - 2 M v_{2}$ , or $v_{2} = - \frac{1}{2} v_{1}$ . The magnitude of $v_{1}$ is $V$ . Therefore, the magnitude of $v_{2}$ is $∣ v_{2} ∣ = - \frac{1}{2} v_{1} = \frac{1}{2} ∣ v_{1} ∣ = \frac{V}{2}$ .

The angle of $6 0^{\circ}$ is given for the first piece, and the second piece will move in the opposite direction to conserve momentum, but its speed is simply $V /2$ .

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