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**Short Answer Questions**

**Q.1. A boy of mass 50 kg running at 5 m s ^{-1} jumps on to a 20 kg trolley travelling in the same direction at 1.5 m s^{-1}. Find their common velocity. [CBSE 2016]**Here mass of boy m

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If their common velocity be v, then from law of conservation of momentum, we have (m

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The magnitude of action and reaction forces are exactly equal but their directions are mutually opposite.

Total momentum of the gun and the bullet system before firing is zero.

**Recoil: Total Momentum = 0**

**Q.3. The velocity-time graph of a car of 1000 kg mass is given alongside. From the graph answer the following: [CBSE 2010, 2016](а) When is the maximum force acting on the car? Why?(b) What is the retarding force?(c) For how long is there no force acting?Ans.** (a) In the v-t graph, the OA part represents uniformly accelerated motion with an acceleration

∴ Force in this region is maximum having a value

F = ma = 1000 x 3.75 = 3750 N

(b) In the v-t graph the part BC represents uniformly retarded motion with an acceleration.

∴ Retarding force F = ma = 1000 x (-7.5) = - 7500 N

(c) The region AB of the graph represents uniform motion with a constant velocity of 15 m s

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∴

∴ Force exerted on the bullet by the rifle.

F = ma = 0.01 x (1 x 10

(a) initial and final momentum

(b) change of momentum

(c) rate of change of momentum.

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Final momentum remains unchanged at P

(b) Change of momentum of ball remains unchanged at

(c) Initially rate of change of momentum but on reducing time to half, the new rate of change of momentum will be

(i) kg m s

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Mass of an object is a measure of its inertia.

The rock of same size has more inertia than a football because mass of rock is much more.

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Mass of trolley = m

Initial velocity of boy = u

Initial velocity of trolley = u

Final velocity = v

= 4 ms

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Given: F = 2N, t = - 10 s, u = 2 ms

v = 5 ms

Since,

or

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Acceleration remains constant. Thus force also remains constant (as F = ma).

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(Take g = 9.8 m/s

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Height of fall, h = 19.6 m

g = 9.8 ms

Final velocity,

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(a) A footballer kicks a ball, which rolls on the ground and after covering some distance comes to rest.

(b) Only the carrom coin at the bottom of a pile is removed when a fast moving striker hits it. [CBSE 2015]

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(b) The carrom coin at the bottom of a pile comes in a state of motion due to force exerted by the striker on it. However, other coins of pile remain intact due to their inertia of rest.

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**Short Answer Questions****Q.1. (a) State the reason why a bullet of small mass fired from a gun kills a person.(b) A bullet of mass 4 g fired with a velocity of 50 m s ^{-1} enters a wall up to a depth of 10 cm. Calculate the average resistance offered by the wall.(c) How will the depth of penetration into the wall change if a bullet of mass 5 g strikes against it with a velocity of 40 m s^{-1}? Give reason to justify your answer. [CBSE 2016]Ans. **(a) Although mass of a bullet is small but its momentum is very large on account of its extremely large velocity. When a bullet hits a person this large momentum is transferred to the person within a very short time and thus produces a large force and may kill the person.

(b) Here mass of bullet m = 4 g = 0.004 kg, initial velocity u = 50 m s

As per relation v

Average resistance force offered by wall F = ma = 0.004 x (- 12500) = - 50 N

(c) Depth of penetration will decrease because initial velocity of bullet is less but magnitude of retardation due to the wall remains unchanged.

(i) weight of the box is increased,

(ii) the surface on which the box is placed is made more rough? [CBSE 2016]

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(i) Its weight ‘W acting vertically downward, (ii) reaction force 'R’ due to horizontal, surface, (iii) force of push 'F’ and (iv) frictional force 'f’. Forces have been shown in Fig. 9.13.

The box will start sliding on the surface only if the pushing force F is greater than the frictional force f i.e.,F>f.

(i) If weight of the box is increased, the magnitude of applied force must increase.

(ii) If the surface is made more rough, the magnitude of applied force must increase.

(ii) An object of mass 50 kg is accelerated uniformly from a velocity of 4 m/s to 8 m/s in 8 s. Calculate the initial and final momentum of the object. Also find the magnitude of the force exerted on the object. [CBSE 2015]

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SI unit: kg ms

mass, m = 50 kg, velocities,

u = 4 ms

Initial momentum, p

= 200 kg m/s

Final momentum, p

= 400 kg ms

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