Question

A 2000 kg car travelling at 72 km/h run into a concrete wall and stopped in 0.05s. What magnitude of impulse did the wall exert on the car ?​

Answer 1

Answer:

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Explanation:

Given,

$m = 2000 \: kg \\ u = 72 \: km \: {hr}^{ - 1} \\ = 20 \: m {s}^{ - 1} \\ v = 0 \: m {s}^{ - 1} \\ a = \frac{v - u}{t} \\ = \frac{ - 20}{0.05} \\ = - 400 \: m {s}^{ - 2} \\ \\ so \: \:F = ma \\ = 2000 \times - 400 \: kg \: m \: {s}^{ - 2} \\ = - 800000 \: N$

So, pressure exerted by wall on car= 800000 N

Answer 2

Answer:

A 2000 kg car traveling at 72 km/h run into a concrete wall and stopped in 0.05s. The magnitude of impulse did the wall exert on the car will be $40,000\ Ns$

Explanation:

Newton's second law of motion is given by the expression,

$F = ma$                   ...(1)

Where

F - Force

m - Mass of the body

a  - Acceleration due to gravity

'a' is the ratio of velocity and time

That is,   $a = \frac{v}{t}$        ...(2)

Substitute equation(2) in (1)           

$F = m(\frac{v}{t})$                  ...(3)

The impulse  (J) on a body is given by,

$Impulse (J) = Force (F) \times Time (t)$

Or $J = Ft$                 ...(4)

       

From the question,

$m = 2000\ kg$

$v = 72\ km/h = 72\times \frac{5}{18} = 20\ m/s$

$t = 0.05\ s$

Substitute these values into equation (3)

$F= 2000\times \frac{20}{0.05} = 800000\ N$

Thus, force  =  $800000\ N$

Substitute all these values into equation (4)

$J = 800000\times 0.05 = 40000 \ Ns$

The magnitude of impulse  =  $= 40000\ Ns$