A 5.38-g bullet is moving horizontally with a velocity of +351 m/s, where the si
ID: 2260335 • Letter: A
Question
A 5.38-g bullet is moving horizontally with a velocity of +351 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 1214 g, and its velocity is +0.664 m/s after the bullet passes through it. The mass of the second block is 1571 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.
A 5.38-g bullet is moving horizontally with a velocity of +351 m/s, where the sign + indicates that it is moving to the right (see
Explanation / Answer
a)First collision:
momentum is conserved:
initial momentum of bullet = mv = 0.00538*351 = 1.888 kgm/s
momentum of the first block after collision:
= mv = 1.214 * 0.664 = 0.8061 kgm/s
momentum of bullet after this collision = 1.888 - 0.8061 = 1.0819 kgm/s.
Velocity of the bullet after collision = momentum/mass
v = 1.0819/0.00538 = 201.097 m/s
second collision
the velocity of the second block after the inelastic collision is
v = m1v1/(m1 + m2)
v = 0.00538*201.097/(1.571 + 0.00538)
v = 1.082/1.576
v = 0.687 m/s
b)total kinetic energy after the collision
= 0.5* 1.214 * 0.664^2 + 0.5 *(1.571 + 0.00538) * 0.687^2
= 0.268 + 0.372
= 0.64
total kinetic energy before the collision
= 0.5 *0.00538 * 351^2
= 331.41
ratio of the total kinetic energy after the collision to that before the collision
= 0.64 / 331.41
= 1.93*10^-3
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