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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.
- 2 m/s
- 4 m/s
- 3 m/s
- 6 m/s
Correct answer: 2 m/s
Solution
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.
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