Question

An 8gm bullet is fired horizontally into a 9kg block of wood and sticks in it. The block which is free to move, has a velocity of 40 cm/s after impact. The initial velocity of the bullet is
A) 450 m/s
B) 450cm/s
C) 220m/s
D) 220m/s

Answer 1

Hint:Here the total net external force acting on the system is zero, that means it is conserved i.e. we can apply conservation of momentum. The formula for conservation of momentum is${m_1}{v_1} = {m_2}{v_2}$. Here we have to tweak the formula here , since there are two masses involved and two velocities are used as well , so the formula for conservation of momentum will become: (${m_1}{u_1} + {m_2}{u_2} = ({m_1} + {m_2})v$). Here ${m_1}$= mass of object 1; ${m_2}$= mass of object 2; ${u_1}$= initial velocity of the object; ${u_2}$= initial velocity of the object 2; v = final velocity.

Complete step-by-step answer:
 Apply conservation of momentum, put values and solve:
${m_1}{u_1} + {m_2}{u_2} = ({m_1} + {m_2})v$;
Put value,( 1kg = 1000gms, 9kg = 9000gms)
$8 \times {u_1} + 9000gms \times {u_2} = (8 + 9000gms) \times 40$;
Here, initial velocity of the block is zero, ${u_2} = 0$,
$8 \times {u_1} + 9000gms \times 0 = (8 + 9000gms) \times 40$;
Solve and find the initial velocity of the bullet
$8 \times {u_1} = (8 + 9000gms) \times 40$;
The initial velocity of the bullet is:
${u_1} = \dfrac{{(8 + 9000gms) \times 40}}{8}$;
${u_1} = 450m/s$

Final Answer:Here the initial velocity of the bullet is 450m/s.

Note:Here the final velocity is the same for the bullet and for the wooden block, this is so because the bullet is stuck in the wooden block and the whole mass is moving with the velocity of 40m/s. Make sure to convert the kg unit into grams and then convert back to kg. Here first use CGS (Centimeter, Gram, Second) system and then convert it into MKS (Meter, Kilogram, Second) system.