A block of mass m = 2 kg has a speed V and is behind a block ofmass M = 44 kg th

Question

A block of mass m = 2 kg has a speed V and is behind a block ofmass M = 44 kg that has a speed of 0.5 m/s. The surface isfrictionless. The blocks collide and couple. After the collision,the blocks have a common speed of 0.9 m/s. In Fig. 8.2, the impulseon the 2-kg block due to the collision is closest to:

0.80 {N}cdot{s}
40 {N}cdot{s}
22 {N}cdot{s}
31 {N}cdot{s}
18 {N}cdot{s}

Now i have worked the math and got that the speed V was 9.7 butwhen it comes time to calculate the impulse which is p2-p1 i don'tknow if it is the momentum of the two objects together or is itonly of the 2kg object.

i calculated 41.4-19.4 and got 22N but that turns out to be thewrong answer so then i did 1.8-19.4 and got -17.6 but that is noteven one of the choices i have been given. i am pretty sure i amjust doing some stupid mistake can somebody help me.

Explanation / Answer

m = 2 kg, v = unknown M = 44 kg, V = 0.5 m/s u = 0.9 m/s impulse on m = -(impulse on M) = -(Mu - MV) = M(V - u) = 44*(0.5 -0.9) = -17.6 Ns magnitude = 18 Ns

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