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slega [8]
3 years ago
6

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 sp

eed 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.
Physics
1 answer:
charle [14.2K]3 years ago
6 0

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 %

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