Question

A car crashes against a wall. The initial velocity at collision is 15 m/sec and the velocity after collision is 2.6 m/sec in the opposite direction. The mass of the car is 1500 kg. What is the average force exerted on the automobile bumper if collision lasts for 0.15 seconds.

1.

1.76 × 10⁵ N

2.

2.1 × 10⁵ N

3.

2.76 × 10⁵ N

4. None of these

Answer 1

Correct Answer - Option 1 :

1.76 × 10⁵ N

Concept:

Impulse (I): It is defined as the integral of force with respect to time. It is a vector quantity. Or it is also defined as a change in the linear moment (P) with respect to time.

I = F  × dt = ΔP 

According to Newton's second law, the Force can be defined as a moment per time.

∴  \({\bf{F}} = \frac{{{\bf{d}}\left( {{\bf{mv}}} \right)}}{{{\bf{dt}}}}\)

This is known as the Impulse momentum equation.

where, m = mass of body, v = relative velocity

Calculation:

Given:

m = 1500 kg, v₁ = 15 m/s, v₂ = 2.6 m/s, t = 0.15 s

Therefore, \(F=\frac{1500\times (15-(-2.6))}{0.15}=\frac{1500\times17.6}{0.15}=1.76\times10^5~N\)