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Over [174]
3 years ago
7

A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string h

orizontal. At the lowest point in its swing when it is moving horizontally, the ball collides elastically with a 2.0-kg block initially at rest on a horizontal frictionless surface. What is the speed of the block just after the collision
Physics
1 answer:
hram777 [196]3 years ago
8 0

Answer:

v_{2}=3.5 m/s

Explanation:

Using the conservation of energy we have:

\frac{1}{2}mv^{2}=mgh

Let's solve it for v:

v=\sqrt{2gh}

So the speed at the lowest point is v=7 m/s

Now, using the conservation of momentum we have:

m_{1}v_{1}=m_{2}v_{2}

v_{2}=\frac{1*7}{2}

Therefore the speed of the block after the collision is v_{2}=3.5 m/s

I hope it helps you!

       

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7 0
3 years ago
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Sergeeva-Olga [200]

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