A bullet with a mass of 7.00 g, traveling horizontally with a speed of 400 m/s,

Question

A bullet with a mass of 7.00 g, traveling horizontally with a speed of 400 m/s, is fired into a wooden block with mass of 0.810 kg, initially at rest on a level surface. The bullet passes through the block and emerges with its speed reduced to 180 m/s. The block slides a distance of 44.0 cm along the surface from its initial position.

a)What is the coefficient of kinetic friction between block and surface?

this part is really buzzing me , I do not know what equation to use !

b)What is the decrease in kinetic energy of the bullet?

I solve this part and the answer is 447J

c)What is the kinetic energy of the block at the instant after the bullet passes through it? 

thanks !

Explanation / Answer

mass of bullet m = 7 g = 7 * 10 ^ -3 kg Initial speed of the bullet u = 400 m / s mass of block M = 0.81 kg Initial speed of the block U = 0 Speed of the bullet after emerging from block v = 180 m / s sliding distance of the block S = 44 cm =0.44 m from law of conservation of momentum , mu + MU = mv + MV 2.8 + 0 = 1.26 + 0.81 V from this speed of the block just after collision V = 1.901 m / s final velocity of the block V ' = 0 from the relation V'^ 2- V^ 2 = 2aS accleration a = [ V'^ 2- V^ 2 ] / 2S = -4.107 m / s^ 2 we know a = -uk* g from this the coefficient of kinetic friction between block and surface uk = -4.107 / g = 0.419 (c). the kinetic energy of the block at the instant after the bullet passes through it is = ( 1/ 2) MV^ 2 = 1.4635 J

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