Question

A thin block of soft wood with a mass of 0.072 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is fired with a speed of 629 m/s at a block of wood and passes completely through it. The speed of the block is 17 m/s immediately after the bullet exits the block.(a) Determine the speed (in m/s) of the bullet as it exits the block. m/s.(b) Determine if the final kinetic energy of this system (block and bullet) is equal to, less than, or greater than the initial kinetic energy. equal to the initial kinetic energy less than the initial kinetic energy greater than the initial kinetic energy.(c) Verify your answer to part (b) by calculating the initial and final kinetic energies of the system in joules. KEi = J KEf = J.

Answer 1

Answer:

366.90149 m/s

923.821735 J

324.734 J

Initial Kinetic energy > Final kinetic energy

Explanation:

$m_1$ = Mass of block = 0.072 kg

$m_2$ = Mass of bullet = 4.67 g

$u_1$ = Initial Velocity of block = 0

$u_2$ = Initial Velocity of bullet = 629 m/s

$v_1$ = Final Velocity of block = 17 m/s

$v_2$ = Final Velocity of bullet

In this system the linear momentum is conserved

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow v_2=\frac{m_{1}u_{1}+m_{2}u_{2}-m_1v_1}{m_2}\\\Rightarrow v_2=\frac{0.072\times 0+4.67\times 10^{-3}\times 629-0.072\times 17}{4.67\times 10^{-3}}\\\Rightarrow v_2=366.90149\ m/s$

Final Velocity of bullet is 366.90149 m/s

The initial kinetic energy

$K_i=\frac{1}{2}m_2u_2^2\\\Rightarrow K_i=\frac{1}{2}4.67\times 10^{-3}\times 629^2\\\Rightarrow K_i=923.821735\ J$

The final kinetic energy

$K_f=\frac{1}{2}m_2v_2^2+\frac{1}{2}m_1v_1^2\\\Rightarrow K_f=\frac{1}{2}4.67\times 10^{-3}\times 366.90149^2+\frac{1}{2}0.072\times 17^2\\\Rightarrow K_f=324.734\ J$

Initial Kinetic energy > Final kinetic energy

Answer 2

To solve the problem we must know about the concept of the law of conservation of momentum.

What is the law of conservation of momentum?

According to the law of conservation of momentum, the momentum of a system is always conserved, therefore, the sum of the initial momentum is equal to the sum of the final momentum.

Given to us

The mass of the bullet, m = 4.67 g= 0.00467 kg

The mass of the wooden block, M = 0.072 = kg

Initial velocity of the bullet, $u_b$ = 629 m/s

Initial velocity of the wooden block, $u_A$ = 0 m/s (rest),

The final velocity of the wooden block, $v_A$ = 17 m/s

A.) We know about the law of conservation of momentum, therefore,

$Mu_A + mu_b = Mv_A + mv_b$

Substitute the values,

$(0.072 \times 0)+ (0.00467 \times 629)= (0.072 \times 17) + (0.00467 \times v_b)$

$v_b$ = 366.9 m/s

Thus, the final velocity of the bullet is 366.9 m/s.

B.) The initial kinetic energy of the system,

The initial kinetic energy of the system can be found by adding the kinetic energy of the wooden block and the kinetic energy of the bullet.

$KE_i = (KE_A)_i + (KE_b)_i$

But since the block was at rest the kinetic energy of the block will be 0 at the initial phase,

$KE_i = (KE_A)_i + (KE_b)_i\\\\KE_i = (KE_b)_i\\\\KE_i = \dfrac{1}{2}m_bu_b^2$

Substitute the values,

$KE_i = \dfrac{1}{2}\times 0.00467 \times 629^2\\\\KE_i = 923.8217\rm\ J$

The final kinetic energy of the system,

The final kinetic energy of the system can be found by adding the kinetic energy of the wooden block and the kinetic energy of the bullet.

$KE_f = (KE_A)_f + (KE_b)_f\\\\KE_f = (\dfrac{1}{2}\times M \times v_A^2)_f + (\dfrac{1}{2}\times m \times v_b^2)_f$

Substitute the values,

$KE_f = (\dfrac{1}{2}\times 0.072 \times 17^2)_f + (\dfrac{1}{2}\times 0.00467 \times 366.9^2)_f\\\\KE_f = 10.404+314.3274\\\\KE_f = 324.734\rm\ J$

Hence, the initial kinetic energy of the system is more than the final kinetic energy of the system.

Learn more about the Law of conservation of momentum:

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