A 5.25kg chunk of ice is sliding at 15.0m/s on the floor of an ice-covered valle

Question

A 5.25kg chunk of ice is sliding at 15.0m/s on the floor of an ice-covered valley when it collides with and sticks to another 5.25kg chunk of ice that is initially at rest. Since the valley is icy, there is no friction.

After the collision, how high above the valley floor will the combined chunks go? (Hint: Break this problem into two parts-the collision and the behavior after the collision-and apply the appropriate conservation law to each part.)

H = ______ m

Please Explain how you found the answers ! Thank You

A 5.25kg chunk of ice is sliding at 15.0m/s on the floor of an ice-covered valley when it collides with and sticks to another 5.25kg chunk of ice that is initially at rest. Since the valley is icy, there is no friction. After the collision, how high above the valley floor will the combined chunks go? (Hint: Break this problem into two parts-the collision and the behavior after the collision-and apply the appropriate conservation law to each part.) H = ______ m Please Explain how you found the answers ! Thank You

Explanation / Answer

For first part:

From the law of conservation of momentum, we have

            m1u1 + 0 = (m1 + m2) v

The combined velocity of the system of two masses is,

            v = m1u1 / m1 + m2

In this case, we have m1 = m2 = m, so the above equation becomes as,

            v = mu1 / 2m

   = u1 / 2

               = 15.0 m/s / 2

                = 7.5 m/s

For second part:

From the law of conservation of energy, we have

   (1/2)2mv2 = 2mgH

Hence, the required height is,

H = v2 /2 g

                           = (7.5 m/s)2 / 2(9.8 m/s2)

                           = 2.87 m

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