Question

A bullet of 10g strikes a wooden block at a speed of 500m/s. What is the resistive force exerted by the wooden block on the bullet if the time taken by the bullet to come in rest is 0.01sec?
1. 5000 N
2. 500 N
3. 50 N
4. None of the above

Answer 1

Correct Answer - Option 2 : 500 N

CONCEPT:

• Momentum: A property of a body in motion that is equal to the product of the body's mass and velocity is called momentum.

P = mv

where P is the momentum of the body, m is the mass of the body, and v is the velocity of the body.

• Newton's Second Law of Motion: It says that the net external force on a system or body is equal to the change in momentum of the system or body divided by the time over which it changes. 

Mathematically:

\(F_{ext} = {Δ p \over Δ t}\)

where F_ext is the external force on the system, Δp is the change in momentum, and Δt is the change in time.

CALCULATION:

Given that m = 10g = 10 × 10^-3 Kg; Δt = 0.01 sec; initial velocity u = 500 m/s; final velocity v = 0.

\(F_{ext} = {Δ p \over Δ t}\)

\(F_{ext} = {mv -mu\over Δ t}\)

\(F_{ext} = {10 \times 10^{-3}\times 500 -0\over 0.01}\)

\(F_{ext} = 500 N\)

So the correct answer is option 2.