Question

A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km h-1. Its velocity is slowed down to 18 km h-7 in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of force required.​

Answer 1

Answer:

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)

Answer 2

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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)

Thanks...