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
A) 21 J
B) 14 J
C) 10 J
D) 0.0012 J
E) 0.16 J

Question

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 
A) 21 J 
B) 14 J 
C) 10 J 
D) 0.0012 J 
E) 0.16 J

Quizplus · Accepted Answer

We can use conservation of energy to solve for the initial kinetic energy of the bullet. The bullet comes to rest in the block, so all of its initial kinetic energy is converted into work done on the block as it lifts upward. We can express this as:

Initial kinetic energy of bullet = work done on block
(1/2)mv^2 = Fd

where m is the mass of the bullet, v is its initial velocity, F is the average force exerted on it during penetration, and d is the distance the block lifts upward.

We can also express F in terms of the acceleration of the bullet during penetration:

F = ma

where a is the acceleration of the bullet.

From the figure, we can see that the block lifts upward 8.0 mm in a time interval of 0.0010 s. The average velocity of the block during this time is:

v_avg = d/t = 0.008 m / 0.001 s = 8 m/s

Since the bullet comes to rest in the block, its final velocity is zero. We can use the kinematic equation:

v_final^2 = v_initial^2 + 2ad

to solve for the acceleration a:

0 = v_initial^2 + 2ad

a = -v_initial^2 / (2d)

The negative sign indicates that the acceleration is opposite to the initial velocity of the bullet.

Substituting this expression for a into our equation relating initial kinetic energy to work done on the block, and solving for the initial kinetic energy, we get:

(1/2)mv^2 = Fd = ma*d = (-v_initial^2 / (2d)) * d * m
v_initial^2 = -2 * (1/2)mv^2 = -mv^2
Initial kinetic energy of bullet = (1/2)mv^2 = -(1/2) * (-mv^2) = (1/2) * m * v^2

We can solve for v using the conservation of momentum:

m_bullet * v_bullet = (m_bullet + m_block) * v_final
v_bullet = v_final * (m_bullet + m_block) / m_bullet

where m_bullet is the mass of the bullet, m_block is the mass of the block, v_final is the velocity of the block-bullet system after the collision, and v_bullet is the velocity of the bullet before the collision.

We can assume that the block initially had no velocity, and that the system is inelastic (i.e. the block and bullet stick together after the collision). Therefore:

m_bullet * v_bullet = (m_bullet + m_block) * v_final
v_bullet = v_final * (m_bullet + m_block) / m_bullet
(15 g) * v_bullet = (2 kg) * v_final
v_bullet = (2 kg / 0.015 kg) * v_final
v_bullet = 267 m/s

Plugging this value of v into the expression we derived earlier for the initial kinetic energy of the bullet, we get:

Initial kinetic energy of bullet = (1/2)mv^2 = (1/2)(0.015 kg)(267 m/s)^2 = 22.4 J

Rounding this to the nearest whole number, we get an answer of 22 J, which is closest to answer choice A). Therefore, the correct answer is:

Answer: A

A 15-G Bullet Is Shot Vertically into an 2-Kg Block