Did anyone have a doubt with the 4th question(additional exersise) NCERT solution of ch-9-Force and laws of motion????
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No, I didn'the have any doubt
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 = − 5 m/s2
Negative sign indicates that its a retarding motion i.e. velocity is decreasing.
Change in momentum = mv − mu = m (v−u)
= 1200 (5 − 25) = − 24000 kg m s−1
Force = Mass × Acceleration
= 1200 × − 5 = − 6000 N
Acceleration of the motor car = − 5 m/s2
Change in momentum of the motor car = − 24000 kg m s−1
Hence, the force required to decrease the velocity is 6000 N.
(Negative sign indicates retardation, decrease in momentum and retarding force)
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