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Q1. Accelerating a 2 kg mass at 5 m/s² requires a force of:
Using Newton’s Second Law, F = ma. Here, mass m = 2 kg and acceleration a = 5 m/s². So F = 2 × 5 = 10 N. This is the force required to produce the given acceleration.
Q2. A motorcar moves at 108 km/h. In m/s, this is:
To convert km/h to m/s, multiply by 5/18. So 108 × 5/18 = 108 ÷ 18 × 5 = 6 × 5 = 30 m/s.
Q3. If a car stops from 30 m/s in 4 seconds, the acceleration is:
Acceleration a = (v – u)/t. Here, initial velocity u = 30 m/s, final velocity v = 0 m/s, time t = 4 s. So a = (0 – 30)/4 = -30/4 = -7.5 m/s². The negative sign indicates deceleration.
Q4. The force applied by the brakes on a 1000 kg car stopping from 30 m/s in 4 s is:
Using F = ma. Mass m = 1000 kg, acceleration a = -7.5 m/s² (from previous question). So F = 1000 × (-7.5) = -7500 N. The negative sign shows force is opposite to motion.
Q5. The negative sign in brake force indicates:
The negative sign in physics indicates direction. For braking force, it is negative because it acts in the opposite direction to the motion of the car, slowing it down.
Q6. A 5 N force gives a mass an acceleration of 10 m/s². The mass is:
Using m = F/a. Here F = 5 N, a = 10 m/s². So m = 5/10 = 0.50 kg.
Q7. A 5 N force gives a mass an acceleration of 20 m/s². The mass is:
Using m = F/a. Here F = 5 N, a = 20 m/s². So m = 5/20 = 0.25 kg.
Q8. When both masses (0.50 kg and 0.25 kg) are tied together, total mass is:
Total mass is the sum of individual masses: 0.50 kg + 0.25 kg = 0.75 kg.
Q9. The acceleration produced by a 5 N force on a 0.75 kg mass is:
Using a = F/m. Here F = 5 N, m = 0.75 kg. So a = 5/0.75 = 6.67 m/s².
Q10. A ball rolling on a level surface slows down from an initial speed to rest. If it decelerates uniformly at 2 cm/s² and takes 10 seconds to stop, its initial velocity was:
Using v = u + at. Here v = 0, a = -2 cm/s², t = 10 s. So 0 = u + (-2)(10) => u = 20 cm/s.
Q11. For the ball in the previous scenario, the time taken to come to rest is:
The time taken is given in the question itself: the ball takes 10 seconds to stop.
Q12. The acceleration of the ball in the scenario is:
The acceleration is -2 cm/s². Converting to m/s²: 2 cm/s² = 2/100 = 0.02 m/s². So both -2 cm/s² and -0.02 m/s² are correct.
Q13. The mass of an object is 20 g. What is its mass in kilograms?
To convert grams to kilograms, divide by 1000. So 20 g = 20/1000 = 0.02 kg.
Q14. The force exerted by the table on the 20 g ball (which decelerates at 0.02 m/s²) is:
Mass = 20 g = 0.02 kg, acceleration = -0.02 m/s². Using F = ma = 0.02 × (-0.02) = -0.0004 N.
Q15. The negative sign of the force on the ball means:
The negative sign indicates that the force is acting in the opposite direction to the ball’s motion, which is why the ball is slowing down.
Q16. Newton’s third law states:
Newton’s Third Law states that for every action force, there is an equal and opposite reaction force. These forces act on different objects.
Q17. Action and reaction forces act on:
Action and reaction forces always act on different objects. If object A exerts a force on object B, then object B exerts an equal and opposite force on object A.
Q18. In football, when two players collide:
When two players collide, each player exerts a force on the other. According to Newton’s Third Law, these forces are equal in magnitude and opposite in direction.
Q19. When spring balances A and B are pulled, readings are:
According to Newton’s Third Law, the force exerted by A on B is equal and opposite to the force exerted by B on A. So both spring balances show the same reading.
Q20. In walking, we push the ground:
When we walk, we push the ground backward with our feet. The ground then pushes us forward with an equal and opposite force, allowing us to move forward.
Q21. The ground pushes us:
According to Newton’s Third Law, when we push the ground backward, the ground pushes us forward with an equal and opposite force. This is what makes us move forward.
Q22. A gun recoils because:
When a gun is fired, the bullet is pushed forward. The bullet exerts an equal and opposite force on the gun, causing it to recoil backward. This is Newton’s Third Law.
Q23. The recoil of the gun is less because:
Recoil velocity = (mass of bullet × velocity of bullet) / mass of gun. Since the gun has much larger mass than the bullet, its recoil velocity is much smaller.
Q24. When a sailor jumps forward from a boat, the boat moves:
When the sailor jumps forward, he pushes the boat backward with an equal and opposite force (Newton’s Third Law). So the boat moves backward.
Q25. Consider two balls, A and B, colliding. The momentum of ball A before collision is:
Momentum before collision is mass × initial velocity. For ball A, this is mA × uA, where uA is its initial velocity.
Q26. Momentum after collision of ball B is:
Momentum after collision is mass × final velocity. For ball B, this is mB × vB, where vB is its final velocity after collision.
Q27. Rate of change of momentum of A is:
Rate of change of momentum = (final momentum – initial momentum) / time = (mA vA – mA uA)/t = mA (vA – uA)/t.
Q28. During collision, ball A exerts a force on ball B equal to:
According to Newton’s Third Law, the force exerted by A on B is equal and opposite to the force exerted by B on A.
Q29. These forces are called:
The forces that two objects exert on each other during collision are action and reaction forces. They are equal in magnitude and opposite in direction.
Q30. Action and reaction forces always occur:
Action and reaction forces always occur at the same time. They are simultaneous and cannot exist without each other.
Q31. Reaction force arises because:
Reaction force is a consequence of Newton’s Third Law, which states that every action has an equal and opposite reaction.
Q32. In the spring balance setup, force applied by A on B is:
According to Newton’s Third Law, the force applied by spring balance A on B is equal and opposite to the force applied by B on A.
Q33. When the ball slows down uniformly, its velocity-time graph is:
For uniform deceleration, the velocity changes at a constant rate. So the velocity-time graph is a straight line with a negative slope.
Q34. In a calculation for braking force, a negative sign is assigned to the force. What does this signify?
The negative sign indicates that the braking force acts in the opposite direction to the motion of the vehicle, which is why it slows down.
Q35. Two blocks of mass 0.50 kg and 0.25 kg are connected together. What is their total mass?
Total mass is the sum: 0.50 kg + 0.25 kg = 0.75 kg.
Q36. Acceleration of combined mass is:
Using a = F/m. With F = 5 N and m = 0.75 kg, a = 5/0.75 = 6.67 m/s².
Q37. Momentum before collision of A is:
Momentum before collision is mass × initial velocity. For ball A, this is mA × uA.
Q38. During collision, forces between A and B are:
According to Newton’s Third Law, the forces that two objects exert on each other during collision are always equal in magnitude and opposite in direction.
Q39. The quantity m(v – u)/t represents:
m(v – u)/t = m × (change in velocity/time) = m × a = F. This is Newton’s Second Law expressed in terms of change in momentum.
Q40. Decrease in velocity to zero means:
When velocity decreases to zero, the acceleration is negative (deceleration). The object is slowing down.
Q41. When mass is doubled and force same, acceleration:
From a = F/m, if mass doubles and force is constant, acceleration becomes half (inversely proportional).
Q42. The boat moves backward when sailor jumps forward due to:
When the sailor jumps forward, he pushes the boat backward. This is due to action-reaction forces (Newton’s Third Law).
Q43. A gun moves backward because:
The gun recoils because the action force on the bullet is accompanied by an equal and opposite reaction force on the gun.
Q44. When brakes act, force is:
Braking force acts in the direction opposite to the motion of the vehicle to slow it down.
Q45. In velocity-time graph, area under graph gives:
The area under a velocity-time graph represents the displacement (or distance) traveled by the object.
Q46. The force causing deceleration of the ball on the table is:
The ball slows down due to the force of friction between the ball and the table surface. Friction opposes motion.
Q47. 20 g in kg is:
To convert grams to kilograms, divide by 1000. So 20 g = 20/1000 = 0.02 kg.
Q48. Force of -0.0004 N means:
The negative sign indicates that the force acts in the opposite direction to the motion, opposing it.
Q49. Newton’s third law pairs are equal in:
Action and reaction forces are equal in magnitude (size) but opposite in direction. They act on different objects.
Q50. In a collision, the forces between objects last for:
During a collision, the forces between objects act for a very short interval of time (t), during which the objects are in contact.
Q51. A cyclist traveling at 6 m/s applies the brakes and comes to a stop in 3 seconds. What is the cyclist’s acceleration?
Using a = (v – u)/t. Here u = 6 m/s, v = 0 m/s, t = 3 s. So a = (0 – 6)/3 = -6/3 = -2 m/s². The negative sign indicates deceleration.