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For a ball moving on a smooth frictionless surface, choose the appropriate option that will make the following statements physically correct. (i) If no net force is applied on the ball, the velocity of the ball will remain the same/increase/decrease. (ii) If a net force is applied on the ball in the direction of its motion, the magnitude of the velocity of the ball will remain the same/increase/decrease. (iii) If a net force is applied on the ball in a direction opposite to the direction of its motion, the magnitude of the velocity of the ball will remain the same/increase/decrease.
In (i), according to Newton's first law, an object continues moving at constant velocity without external force, so it will remain the same. In (ii), a forward net force causes acceleration in the direction of motion, so the magnitude of velocity will increase. In (iii), an opposing net force createRead more
In (i), according to Newton’s first law, an object continues moving at constant velocity without external force, so it will remain the same. In (ii), a forward net force causes acceleration in the direction of motion, so the magnitude of velocity will increase. In (iii), an opposing net force creates negative acceleration opposing motion, so the magnitude of velocity will decrease.
For more NCERT Solutions of Class 9 Science Exploration Chapter 6 How Forces Affect Motion Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-6/
See lessUsing a horizontal force F, a table is moved across the floor at a constant velocity. How much is the frictional force exerted by the floor on the table?
Since the table moves across the floor at a constant velocity, its acceleration is zero. By Newton's first law of motion, the net external horizontal force acting on the table must be zero. Therefore, the opposing frictional force exerted by the floor on the table must be exactly equal in magnitudeRead more
Since the table moves across the floor at a constant velocity, its acceleration is zero. By Newton’s first law of motion, the net external horizontal force acting on the table must be zero. Therefore, the opposing frictional force exerted by the floor on the table must be exactly equal in magnitude to the applied horizontal force F, acting in the opposite direction.
For more NCERT Solutions of Class 9 Science Exploration Chapter 6 How Forces Affect Motion Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-6/
See lessA coconut of mass 1.5 kg falls from the top of a coconut tree onto the wet sand on a beach. The height of the tree is 10 m. On impact, the coconut comes to rest by making a depression in the sand. (i) Calculate the velocity of the coconut just before it hits the sand. (ii) Assume that the average resistive force of sand is 3000 N and all of the coconut’s energy is used to create the depression in the sand. Calculate the depth of the depression the coconut makes in the sand. Assume g = 10 m s-2.
Using conservation of energy, the velocity before impact equals square root of two times g times h, which is square root of 200, giving 14.14 m s-1. The kinetic energy on impact equals initial potential energy m g h, totaling 150 J. Equating this energy to work done against sand resistance, 150 J diRead more
Using conservation of energy, the velocity before impact equals square root of two times g times h, which is square root of 200, giving 14.14 m s-1. The kinetic energy on impact equals initial potential energy m g h, totaling 150 J. Equating this energy to work done against sand resistance, 150 J divided by 3000 N yields a depression depth of 0.05 m or 5 cm.
For more NCERT Solutions of Class 9 Science Exploration Chapter 7 Work, Energy and Simple Machines Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-6/
See lessThe potential energy-displacement graph of a 0.5 kg ball moving along a frictionless track is shown in Fig. 7.39. At O, the velocity of the ball is 0 m s-1 and potential energy is 30 J. Calculate the velocity of the ball at P, Q and R.
According to the potential energy-displacement graph in Fig. 7.39, total mechanical energy is fixed at 30 J. At point P, the graph indicates potential energy is 20 J, leaving 10 J for kinetic energy, which gives a speed of 6.32 m s-1. At point Q, potential energy is 30 J, meaning kinetic energy is 0Read more
According to the potential energy-displacement graph in Fig. 7.39, total mechanical energy is fixed at 30 J. At point P, the graph indicates potential energy is 20 J, leaving 10 J for kinetic energy, which gives a speed of 6.32 m s-1. At point Q, potential energy is 30 J, meaning kinetic energy is 0 J and velocity is 0 m s-1. The ball cannot reach R.
For more NCERT Solutions of Class 9 Science Exploration Chapter 7 Work, Energy and Simple Machines Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-7/
See lessA 1000 kg car is moving along a road at a constant speed. Suddenly, the driver notices some obstruction ahead and applies the brakes to come to a complete stop. The graphical representation of motion of the car starting from the instant the driver spots the traffic ahead is shown in Fig. 7.38. (i) Describe how the car moves between positions A and B. (ii) Calculate the kinetic energy of the car at A. (iii) State the work done by the brakes in bringing the car to a halt between B and C. (iv) What does the kinetic energy of the car transform into?
Using the speed-time graph in Fig. 7.38, the car moves with uniform speed at 35 m s-1 from time 0 to 1 second between A and B. Its kinetic energy at A is half multiplied by 1000 kg multiplied by 35 squared, equaling 612500 J. Between B and C, the brakes bring speed to zero, doing minus 612500 J of wRead more
Using the speed-time graph in Fig. 7.38, the car moves with uniform speed at 35 m s-1 from time 0 to 1 second between A and B. Its kinetic energy at A is half multiplied by 1000 kg multiplied by 35 squared, equaling 612500 J. Between B and C, the brakes bring speed to zero, doing minus 612500 J of work, dissipating energy as heat and sound.
For more NCERT Solutions of Class 9 Science Exploration Chapter 7 Work, Energy and Simple Machines Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-7/
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