Question

A 20 gram bullet, traveling horizontally with a velocity of magnitude 350 m/s, is fired into a wooden block with a mass of 0.75, initially at rest on a level surface. The bullet passes through the block and emerges with its speed reduced to 150 m/s. What is the kinetic energy of the block (in J) at the instant after the bullet passes through it

Answer 1

Answer:

The kinetic energy is 10.65 joules.

Explanation:

This case can be considered as a collision if we neglect the frictional forces the ball experiences traveling through the wooden block because are lower than the impact forces, then we can use conservation of linear momentum to model this situation considering the system block-bullet.

Linear momentum states total initial momentum (pi) is equal to the final momentum (pf), this is:

$p_f=p_i$

Total momentum is the sum of individual momentums, in our case the momentum of the ball (pb) plus the momentum of the wooden block (pw), then:

$p_{bf}+p_{wf}=p_{bi}+p_{wi}$

Linear momentum of an object is defined as mass times velocity, then:

$m_bv_{bf}+m_wv_{wf}=m_bv_{bi}+m_wv_{wi}$

With vb the velocities of the bullet, mb the mass of the bullet, vw the velocities of the wooden block and mw the mass of the wooden block. Note that the wooden block is initially at rest, that means the initial velocity of the wooden block is zero, then:

$m_bv_{bf}+m_wv_{wf}=m_bv_{bi}$

Solving the equation for the final velocity of the block:

$v_{wf}=\frac{m_bv_{bi} - m_bv_{bf}}{m_w}=\frac{m_b(v_{bi} - v_{bf})}{m_w}$

$v_{wf}=\frac{0.02(350- 150)}{0.75} =5.33 \frac{m}{s}$

(Important: i. The sign of the velocities is important, so we chose positive direction as the direction the bullet moves initially, positive velocities means they are in this same direction.

ii. I assume 0.75 kg for the mass of the wooden block because you didn't provide its unities)

Now with the velocity of the bullet after the collision we can find the kinetic energy using the equation for kinetic energy:

$K=\frac{m_wv_{wf}^2}{2}=\frac{(0.75)(5.33)^2}{2}$

$K=10.65 J$

Answer 2

Answer:

The kinetic energy of the block at the instant after the bullet passes through it is 1000 J

Explanation:

Given:

mass of bullet = m = 20 g = 20x10³ kg

initial velocity = u₁ = 350 m/s

final velocity = u₂ = 150 m/s

The loss in kinetic energy is

$E_{K} =\frac{1}{2} m(u_{2}^{2} -u_{1} ^{2} )=\frac{1}{2} *20x10^{-3} (350^{2} -150^{2} )=1000J$