Question

A 10.0-g bullet is fired into a stationary block of wood having mass 5.00 kg. The bullet embeds 10 pts into the block and the speed of the block-and-bullet after the collision is 0.600 m/s. Find a) the original speed of the bullet, b) the mechanical energies of the block-bullet system before and after the collision, c) the percentage of mechanical energy lost to heat.

Answer 1

Answer:

A. The initial velocity of the bullet is $= 300.6m/s$

B. Mechanical energy of the system before and after collision: 451.80 J, 0.9018 J

C. Percentage of K.E lost to heat is  = 99.8 %

Explanation:

From conservation of linear momentum,

$(m_{1}v_{1} +m_{2}v_{2})= (m_{1}+m_{2})v$

let the mass of the block be m1 and velocity = v1

let the mass of the bullet be m2 and velocity = v2

Let the final velocity of the system be v.

A. Plugging our parameters into the equation, we have:

$[(5 \times 0) +(0.01\times v_{2})]= 5.01 \times 0.6$

$v_{2}=\frac{3.006}{0.01}= 300.6m/s$

Hence, the initial velocity of the bullet is $= 300.6m/s$

B. The mechanical energies of the system exist in form of kinetic energy.

I. Kinetic energy of the system before collision:

$0.5 \times 5\times 0^{2} + 0.5 \times 0.01 \times 300.6^{2}= 451.80 J$

II. Kinetic energy after collision:

$0.5\times 5.01 \times 0.6^{2}= 0.9018 J$

C. Change in Mechanical Energy = $451.8 - 0.9018 J= 450.9J$

$\frac{450.9}{451.8} \times 100 =99.8%$

Percentage of K.E lost to heat is  = 99.8 %