Question

A car of 600kg mass was moving 0.2 m/s^2 accelaration. the car then hits a pickup of 400kg after 60 sec. After collision they move with a velocity of 7.2 m/s^1


a. Find the applied force of first car

Answer 1

Answer:

Explanation:

Given,

m1 = 600kg

a = 2m/s^2

m2 = 400kg

u2 = 0m/s ( since truck was at rest )

Using $m1u1 + m2u2 = m1v1 + m2v2$,

we get  600(u1) + 0 = 4320 + 2880

      ⇒ u1 = (7200)/600

               = 12m/s

v1 = u1 + at = 12 + 12 = 24m/s

We can use the formula,

$F = \frac{m(v-u)}{t}$

⇒ {600(24-12)}/60

  = 120kgm/s^2

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