A car is moving with a uniform velocity of 30 m $s^{- 1}$ . It is stopped in 2 s by applying a force of 1500 N through its brakes. Calculate:

(a) the change in momentum of car,
(b) the retardation produced in car and
(c) the mass of the car.

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Solution

Given:

Initial velocity = 30 m/s

Final velocity = 0 m/s

Time required, t = 2 s

Force applied = -1500N ( here the negative sign indicates the force is applied in the opposite direction of the motion. Therefore it causes deceleration and makes the car come to rest.)

(i) F = $fraction numerator c h a n g e space i n space m o m e n t u m over denominator t i m e end fraction$

Change in momentum = Fx t = -3000 kg m/s

The negative sign tells us that the momentum has decreased.

(ii) Acceleration, a = $fraction numerator v minus u over denominator t end fraction equals fraction numerator 0 minus 30 over denominator 2 end fraction equals negative 15 m s to the power of negative 2 end exponent$

Hence, the retardation is $15 m / s^{2}$ .

Mass of the car = $fraction numerator F o r c e space over denominator a c c e l e r a t i o n end fraction equals fraction numerator negative 1500 over denominator negative 15 end fraction$ = 100 kg

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A car is moving in a straight line with speed 18 km $h^{- 1}$ . It is stopped in 5 s by applying the brakes. Find:

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(ii) the retardation of car
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