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Force and Laws of Motion | Additional Exercises 2

Question 2

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. Calculate the acceleration and change in momentum. Also, calculate the magnitude of the force required.

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Given, mass of the car (m) = 1200kg

Initial velocity (u) = 90 km/hour = 25 meters/sec

Terminal velocity (v) = 18 km/hour = 5 meters/sec

Time period (t) = 4 seconds

\text { The acceleration of the car can be calculated with the help of the formula: } a=\frac{v-u}{t}

a=\frac{5-25}{4} m . s^{-2}=-5 \mathrm{ms}^{-2}

\text { Therefore, the acceleration of the car is }-5 \mathrm{ms}^{-2}

Initial momentum of the car = m × u = (1200kg) × (25m/s) = 30,000 kg.m.s^{-1}

Final momentum of the car = m × v = (1200kg) × (5m/s) = 6,000 kg.m.s^{-1}

Therefore, change in momentum (final momentum – initial momentum) = (6,000 – 30,000) kg.m.s^{-1}

= -24,000 kg.m.s^{-1}

External force applied = mass of car × acceleration

= (1200kg) × (-5 ms^{-2}) = -6000N

Therefore, the magnitude of force required to slow down the vehicle to 18 km/hour is 6000 N

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Question 2

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. Calculate the acceleration and change in momentum. Also, calculate the magnitude of the force required.

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

Given, mass of the car (m) = 1200kg

Initial velocity (u) = 90 km/hour = 25 meters/sec

Terminal velocity (v) = 18 km/hour = 5 meters/sec

Time period (t) = 4 seconds

\text { The acceleration of the car can be calculated with the help of the formula: } a=\frac{v-u}{t}

a=\frac{5-25}{4} m . s^{-2}=-5 \mathrm{ms}^{-2}

\text { Therefore, the acceleration of the car is }-5 \mathrm{ms}^{-2}

Initial momentum of the car = m × u = (1200kg) × (25m/s) = 30,000 kg.m.s^{-1}

Final momentum of the car = m × v = (1200kg) × (5m/s) = 6,000 kg.m.s^{-1}

Therefore, change in momentum (final momentum – initial momentum) = (6,000 – 30,000) kg.m.s^{-1}

= -24,000 kg.m.s^{-1}

External force applied = mass of car × acceleration

= (1200kg) × (-5 ms^{-2}) = -6000N

Therefore, the magnitude of force required to slow down the vehicle to 18 km/hour is 6000 N

Our top 5% students will be awarded a special scholarship to Lido.

subject-cta
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