A 1.03 10-2 kg bullet is fired horizontally into a 2.65 kg wooden block attached

Question

A 1.03 10-2 kg bullet is fired horizontally into a 2.65 kg wooden block attached to one end of a massless, horizontal spring (k = 844 N/m). The other end of the spring is fixed in place, and the spring is unstrained initially. The block rests on a horizontal, frictionless surface. The bullet strikes the block perpendicularly and quickly comes to a halt with in it. As a result of this completely inelastic collision, the spring is compressed along its axis and causes the block/bullet to oscillate with an amplitude of 0.200 m. What is the speed of the bullet?

Explanation / Answer

Since friction acts against the motion, which is uphill so friction will acts towards down side along with mg sin 26.4 in the direction of AB and is given as mu*R (mu = 0.52, R = mg cos 26.4) so Total resistance force acting downwards = mg sin 26.4 + 0.52*mg cos 26.4 where "m" is the mass of the block in kg resistance force = 8.7*9.81*sin 26.4 + 0.52*8.7*9.81 cos 26.4 = 77.7 N since F = ma ; where a = retardation a = F/m = 77.7/8.7 = 8.931 m/sec^2 since we know final vel is zero so v^2 = u^2 - 2as ;u = 1.56 m/s, s = distance, v = 0, a = 8.931 0 = 1.56^2 - 2*8.931*s s = 0.136 m ------------- Answer --------------------------------------… FROM the conservation of energy : KE = PE + work done against friction 1/2 mv^2 = mgh + F*d F = mu* mg cos 26.4 =>0.52*8.7*9.81*cos 26.4 = 39.75 N, d = distance on ramp, h = vertical height 1/2*8.7*(1.56)^2 = 8.7*9.81*{d*sin 26.4} + 39.75*d ; sin 26.4 = h/d so h = d* sin 26.4 10.5861 = 77.6982 d or d = 0.136 meter So both way you get the same answer, so it will slide 0.136 meter before it stops

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