A 2 kg block sits on a horizontal floor. A 1 kg block, connected to it by an inextensible string passing over a pulley, is released from rest and falls 0.9 m before the string suddenly becomes taut. Find the common speed (in m/s) of the two blocks just after the string jerks tight.

Question

Accepted Answer

2 m/s. The falling block reaches sqrt(2*10*0.9) = sqrt(18)... using g=10 gives v0; momentum conservation during the sudden jerk shares this between both masses, giving a common speed of about 2 m/s.

A 2 kg block sits on a horizontal floor. A 1 kg block, connected to it by an inextensible string passing over a pulley, is released from rest and falls 0.9 m before the string suddenly becomes taut. Find the common speed (in m/s) of the two blocks just after the string jerks tight.

Solution

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