A 7.45g bullet is red at a 4.0kg wooden block hanging from a very light string.

Question

A 7.45g bullet is red at a 4.0kg wooden block hanging from a very light string. The bullet strikes the stationary block at 360m/s, and lodges in the wood. The bullet/block combination then swings upward to a height h above its initial position. (a)What is the speed of the bullet/block combination immediately after the collision? (b)What is the coecient of restitution for this collision? Why? (c)Find the center of mass kinetic energy and the initial value of the convertible energy for this collision. (d)How much kinetic energy is actually converted to other forms in this collision? What percentage of the original kinetic energy is that? (e)Where does that energy go? (f)Find the height h to which the bullet/block combination rises after the impact. Assume the total mechanical energy is conserved for this part of the motion, and there is no energy loss or gain associated with the string

Explanation / Answer

Mass of the bullet m1 0.00745 kg mass of the wood block m2 4 kg initial velocity of bullet u1 360 m/s initial velocity of block u2 0 (a) After impact bullet and block travelled together. That is their velocities are same. Thus v1=v2=(m1u1+m2u2)/(m1+m2)m = 0.66925 m/sec (b) Coeffcient of restitution e= v1-v2/(u1-u2) =( 0.66925- 0.66925)/(360-1) =0, as it is a plastic impact where both the masses travel together. ( c ) The center of mass of kinetic energy is at the centre of bullet before the impact. After the collision, the center of mass of kinetic energy is the at the center of wooden block as mass of bullet is very small. (d) Intial kinetic energy of bullet is 0.5mu12 = 482.76 j kinetic energy after impact = 0.5(m1+m2)v12 = 0.897 J thus a very small part of the available energy is converted (0.897/482.76 = 0.0018 i.e less than 0.18% and reaming 99.82% is lost. ( e ) the lost kinetic energy has gone in bullet penetrating the wood. (f) All the available kinetic energy after collision, will be converved in increaing the potentiual energy of the bullet+ wwoden block combination. Thus we have 0.5(m1+m2)v12 = (m1+m2)gh where h is the height the block would go up. thus we have h = 0.5v12/g =0.5 x 0.669252/9.81 = 0.02283 m

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