Question

Did anyone have a doubt with the 4th question(additional exersise) NCERT solution of ch-9-Force and laws of motion????

Answer 1

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)