**Additional Exercises Page: 130** ( Force And Laws Of Motion )

**1. The following is the distance-time table of an object in motion:**

Time (seconds) | Distance (meters) |

0 | 0 |

1 | 1 |

2 | 8 |

3 | 27 |

4 | 64 |

5 | 125 |

6 | 216 |

7 | 343 |

**(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero? (b) What do you infer about the forces acting on the object?**

**Solution**

(a) The distance covered by the object at any time interval is greater than any of the distances covered in previous time intervals. Therefore, the acceleration of the object is increasing.

(b) As per the second law of motion, force = mass × acceleration. Since the mass of the object remains constant, the increasing acceleration implies that the force acting on the object is increasing as well

**2. Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 m s ^{-2}. With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort)**

**Solution**

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

When the third person starts pushing the car, the acceleration (a) is 0.2ms^{-2}. Therefore, the force applied by the third person (F = ma) is given by:

F = 1200kg × 0.2 ms^{-2} = 240N

The force applied by the third person on the car is 240 N. Since all 3 people push with the same muscular effort, the force applied by each person on the car is 240 N.

**3. A hammer of mass 500 g, moving at 50 m s-1, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?**

**Solution**

Given, mass of the hammer (m) = 500g = 0.5kg

Initial velocity of the hammer (u) = 50 m/s

Terminal velocity of the hammer (v) = 0 (the hammer is stopped and reaches a position of rest).

Time period (t) = 0.01s

a = -5000ms^{-2}

Therefore, the force exerted by the hammer on the nail (F = ma) can be calculated as:

F = (0.5kg) * (-5000 ms^{-2}) = -2500 N

As per the third law of motion, the nail exerts an equal and opposite force on the hammer. Since the force exerted on the nail by the hammer is -2500 N, the force exerted on the hammer by the nail will be +2500 N.

**4. 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.**

**Solution**

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

Therefore, the acceleration of the car is -5 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