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Q. A motorcar of mass 1200kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 second by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.

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Answer 1

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Given info : A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force.

To find : the acceleration and change in linear momentum and also magnitude of force required.

solution : initial velocity, u = 90 km/h

= 90 × 5/18

= 25 m/s

final velocity, v = 18 km/h

= 18 × 5/18

= 5 m/s

we know, acceleration = change in velocity with respect to time.

so, acceleration, a = (final velocity - initial velocity)/time taken

= (5 - 25)/4

= -20/4

= -5 m/s²

change in linear momentum = final linear momentum - initial linear momentum

= mv - mu

= m(v - u)

= 1200 kg × (5 m/s - 25 m/s)

= 1200 kg × -20 m/s

= -24000 kgm/s

magnitude of force required, F = ma

= 1200 kg × 5 m/s² [ just taking magnitude of acceleration because we have to find magnitude of force ]

= 6000 N

Answer 2

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A motorcar of mass 1200kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 second by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.

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Mass of the motor car, m = 1200 kg

Initial velocity of the motor car, u = 90 km/h = 25 m/s

Final velocity of the motor car, v = 18 km/h = 5 m/s

Time taken, t = 4 s

According to the first equation of motion:

v = u + at

5 = 25 + a (4)

a = ˆ’ 5 m/s2

Negative sign indicates that its a retarding motion i.e. velocity is decreasing.

Change in momentum = mv ˆ’ mu = m (vˆ’u)

= 1200 (5 ˆ’ 25) = ˆ’ 24000 kg m sˆ’1

Force = Mass — Acceleration

= 1200 — ˆ’ 5 = ˆ’ 6000 N

Acceleration of the motor car = ˆ’ 5 m/s2

Change in momentum of the motor car = ˆ’ 24000 kg m sˆ’1

Hence, the force required to decrease the velocity is 6000 N.

(Negative sign indicates retardation, decrease in momentum and retarding force)