Physics question. I have no idea where to even start.

A 5.55-g bullet is moving horizontally with a velocity of +348 m/s, where the sign + indicates that it is moving to the right (see part a of the drawing). The bullet is approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part b. Note that both blocks are moving after the collision with the bullet. The mass of the first block is 1176 g, and its velocity is +0.580 m/s after the bullet passes through it. The mass of the second block is 1592 g. (a) What is the velocity of the second block after the bullet imbeds itself? (b) Find the ratio of the total kinetic energy after the collision to that before the collision.

image: http://edugen.wileyplus.com/edugen/courses/crs2216/cutnell3550/cutnell3550c07/image_n/nw0297-nu.jpg

*keep in mind that the image has different numbers than my problem

Let m = .00555 kg, M1 = 1.176 kg, M2 = 1.592 kg, v = .58 mps, and U = 348 mps.

From conservation of momentum, we have mU = M1 v + m V; so V = (mU - M1 v)/m = (.00555*348 - 1.176*.58)/.00555 = 225.1027027 = 225.1 mps after the first impact.

Then mV = (m + M2) u for the second impact; so that u = V m/(m + M2) = 225.1*.00555/(.00555+1.592) = 0.782013083 = .782 mps ANS a).

Total preimpact energy is TE = 1/2 mU^2 = (1/2)*.00555*348^2 = 336.0636 = 336.1 J and total kinetic energy after impacts is ke = 1/2 (m + M2) u^2 = (1/2)*(.00555+1.592)*.782^2 = 0.488470083 = .488; so that n = .488/336 = 0.001452381 = .00145 ANS b)

Perhaps we're using the wrong ratio. Maybe they mean the total KE before the second collision rather than the first. Try that ratio.

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Use momentum conservation for each block-bullet step.