At the bottom of its path, the ball strikes a 2.30 kg steel block initially at rest on a frictionless surface. The collision is elastic. Fin

Question

At the bottom of its path, the ball strikes a 2.30 kg steel block initially at rest on a frictionless surface. The collision is elastic. Find (a) the speed of the ball and (b) the speed of the block, both just after the collision.

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Thu Thủy 2 weeks 2021-09-04T20:36:54+00:00 1 Answers 0 views 0

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    2021-09-04T20:38:28+00:00

    Answer:

    (a). The speed of the ball after collision is 2.01 m/s.

    (b). The speed of the block after collision 1.11 m/s.

    Explanation:

    Suppose, A steel ball of mass 0.500 kg is fastened to a cord that is 50.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal.

    Given that,

    Mass of steel block = 2.30 kg

    Mass of ball = 0.500 kg

    Length of cord = 50.0 cm

    We need to calculate the initial speed of the ball

    Using conservation of energy

    \dfrac{1}{2}mv^2=mgl

    v=\sqrt{2gl}

    Put the value into the formula

    u=\sqrt{2\times9.8\times50.0\times10^{-2}}

    u=3.13\ m/s

    The initial speed of the ball u_{1}=3.13\ m/s

    The initial speed of the block u_{2}=0

    (a). We need to calculate the speed of the ball after collision

    Using formula of collision

    v_{1}=(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{1}+(\dfrac{2m_{2}}{m_{1}+m_{2}})u_{2}

    Put the value into the formula

    v_{1}=(\dfrac{0.5-2.30}{0.5+2.30})\times3.13

    v_{1}=-2.01\ m/s

    Negative sign shows the opposite direction of initial direction.

    (b). We need to calculate the speed of the block after collision

    Using formula of collision

    v_{2}=(\dfrac{2m_{1}}{m_{1}+m_{2}})u_{1}+(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{2}

    Put the value into the formula

    v_{2}=(\dfrac{2\times0.5}{0.5+2.30})\times3.13+0

    v_{2}=1.11\ m/s

    Hence, (a). The speed of the ball after collision is 2.01 m/s.

    (b). The speed of the block after collision 1.11 m/s.

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