A man holding a rock sits on a sled that is sliding across a frozen lake (neglig

Question

A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.530 m/s. The total mass of the sled, man, and rock is 93.0 kg. The mass of the rock is 0.310 kg and the man can throw it with a speed of 15.5 m/s. Both speeds are relative to the ground.

Determine the speed of the sled if the man throws the rock forward (i.e. in the direction the sled is moving) in m/s.

Determine the speed of the sled if the man throws the rock directly backward in m/s.

Explanation / Answer

A)

From conservation of momentum...

(m + M)(v) = mv + Mv

(93)(.53) = (.31)(15.5) + (93.5 - .31)(v)

v = .477 m/s

B) Since all the velocities were in 1 direction, I considered all positive.

For backward direction, the v1 of the rock becomes negetive.

So, the equation becomes

m*v = m1*v1 + m2*v2

93*0.53 = -0.31*15.5 + 93*v2

v2 = (93*0.53 + 0.31*15.5)/93 = 0.582 m/s

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