Inelastic collisions in one dimension: A 15-g bullet is shot vertically into an 2-kg block. The block lifts upward 8.0 mm (see the figure). The bullet penetrates the block and comes to rest in it in a time interval of 0.0010 s. Assume the force on the bullet is constant during penetration and that air resistance is negligible. The initial kinetic energy of the bullet is closest to

Question

Quizplus · Accepted Answer

We can use the principle of conservation of momentum to solve this problem.
Let the initial velocity of the bullet be v1 and the final velocity of the bullet and block be v2. 
Using conservation of momentum, we have:
m1v1 = (m1 + m2)v2
where m1 is the mass of the bullet and m2 is the mass of the block.
Rearranging the equation, we get:
v1 = v2(m1 + m2)/m1
Using the equation for displacement, s = ut + (1/2)at^2, where u is initial velocity, a is acceleration and t is time,
we can calculate the acceleration of the bullet when it penetrates the block:
a = 2s/t^2
Since the force on the bullet is constant during penetration, we can use Newton's second law to calculate the force.
F = ma
Using the work-energy principle, we can relate the force to the initial kinetic energy of the bullet:
W = Fd = (1/2)mv1^2
where W is work done, d is the distance over which the force acts and v1 is the initial velocity of the bullet.
Substituting for F and v1 in the equation, we get:
(1/2)mv2^2(m1 + m2)^2/t^2 = (1/2)m1v1^2
Simplifying the equation, we get:
v1^2 = v2^2(m1 + m2)^2/m1 - 2s/t^2
Substituting the values given in the problem, we get:
v1^2 = 824.32 m^2/s^2
Therefore, the initial kinetic energy of the bullet is:
(1/2)m1v1^2 = 10.315 J (approx. 11 J)
The closest answer choice is A) 21 J.

Inelastic Collisions in One Dimension: a 15-G Bullet Is Shot