PHYSICS
11 th
JEE
NEWTON’S LAW OF
MOTION
Aryabhatta Tower 2A-51 (A), Road No.-3, Behind City Mall, Industrial
Area Kota, Rajasthan 324005, Mo. Number 7019243492
, 1
VIDYAPEETH
NEWTON’S LAW OF MOTION
DPP-1
[Motion Momentum, Types of force]
1. A force of 10 N is applied to a mass of 10 gm for 5. A force of 6N acts on a body at rest of mass
10 seconds. The change of momentum in 1 kg. During this time, the body attains a velocity
kgm / sec units will be- of 30 m/s. The time for which the force acts on
the body is –
(A) 10 (B) 100
(A) 10 sec. (B) 8 sec.
(C) 1000 (D) 0.01
(C) 7 sec. (D) 5 sec.
2. A ball moving with a velocity of 20 m/sec has a
6. A bullet is fired from a gun. The force on the
mass of 50gm. It collides against a wall normally
bullet is given by F = 600 − 2 105 t , where F is
and is rebounded normally with the same speed.
in N and t in sec. The force on bullet becomes
If the time of impact of the ball and the wall is 40
zero as soon as it leaves barrel. What is average
milliseconds, the average force exerted on the
impulse imparted to bullet-
wall in dynes is –
(A) 9 Ns (B) zero
7
(A) zero (B) 1.25 × 10
(C) 0.9 Ns (D) 1.8 Ns
7 6
(C) 2.5 × 10 (D) 5 × 10
7. A golf ball of mass 0.05 kg placed on a tee, is
3. A body of mass m is thrown vertically upwards struck by a golf club. The speed of the golf ball
with a velocity v. It reaches to a height h and as it leaves the tee is 100 m/s, the time of contact
returns after t seconds. Then, the total change in between them is 0.02s. If the force decreases to
its momentum is – zero linearly with time, then the force at the
(A) zero (B) mv beginning of the contact is –
(C) mgt (D) mght (A) 500 N (B) 250 N
(C) 200 N (D) 100 N
4. A machine gun fires bullets of 50gm at the speed
of 1000 m/sec. If an average force of 200N is 8. A force F = 6t 2ˆi + 4tˆj , is acting on a particle of
exerted on the gunner, the number of bullets fired mass 3 kg then what will be velocity of particle at
per minute is – t = 3 sec. if at t = 0, particle is at rest-
(A) 240 (B) 120 (A) 18î + 6ˆj (B) 18î + 12ˆj
(C) 60 (D) 30 (C) 12iˆ + 6ˆj (D) None
, 2
9. A cricket player catches a ball of mass 100g 13. A body of mass 2 kg moves with acceleration
moving with a speed of 25 ms–1. If the ball is 3 ms–2. The change in momentum in one second
caught in 0.1s, the average force of the blow is:
2 3
exerted on the hands of the player is (A) kg ms−1 (B) kg ms−1
3 2
(A) 4 N (B) 25 N
(C) 6 kg ms−1 (D) None of these
(C) 40 N (D) 250 N
14. In case of a box lying on a horizontal table
10. The linear momentum P of a particle varies with
(A) Action and reaction are equal and opposite
time as follows, P = + t 2 , where and are and act perpendicular to the surfaces in
constants. The net force acting on the particle is contact
(A) Proportional to t (B) Proportional to t2 (B) Action and reaction are equal but act in the
(C) Zero (D) Constant same direction
(C) Action and reaction are not equal but are in
opposite directions
11. A machine gun fires n bullets per second and the
(D) Action of box on table and reaction of table
mass of each bullet is m. If v is the speed of each
on box are equal and opposite and are
bullet, then the average force exerted on the
inclined to vertical
machine gun is
(A) mng (B) mnv
15. The displacement (x) – time (t) curve of a particle
(C) mnvg (D) (mnv)/g
is shown in figure. The external force acting on
the particle is:
12. What is the impulse of force shown in the
Displacement ( x)
following figure?
F(N)
100 time (t)
O
(A) Acting at the beginning part of motion
t(s)
1 10 (B) Zero
(A) 225Ns (B) 450Ns (C) Not zero
(C) 900Ns (D) 1000Ns (D) None of these
11 th
JEE
NEWTON’S LAW OF
MOTION
Aryabhatta Tower 2A-51 (A), Road No.-3, Behind City Mall, Industrial
Area Kota, Rajasthan 324005, Mo. Number 7019243492
, 1
VIDYAPEETH
NEWTON’S LAW OF MOTION
DPP-1
[Motion Momentum, Types of force]
1. A force of 10 N is applied to a mass of 10 gm for 5. A force of 6N acts on a body at rest of mass
10 seconds. The change of momentum in 1 kg. During this time, the body attains a velocity
kgm / sec units will be- of 30 m/s. The time for which the force acts on
the body is –
(A) 10 (B) 100
(A) 10 sec. (B) 8 sec.
(C) 1000 (D) 0.01
(C) 7 sec. (D) 5 sec.
2. A ball moving with a velocity of 20 m/sec has a
6. A bullet is fired from a gun. The force on the
mass of 50gm. It collides against a wall normally
bullet is given by F = 600 − 2 105 t , where F is
and is rebounded normally with the same speed.
in N and t in sec. The force on bullet becomes
If the time of impact of the ball and the wall is 40
zero as soon as it leaves barrel. What is average
milliseconds, the average force exerted on the
impulse imparted to bullet-
wall in dynes is –
(A) 9 Ns (B) zero
7
(A) zero (B) 1.25 × 10
(C) 0.9 Ns (D) 1.8 Ns
7 6
(C) 2.5 × 10 (D) 5 × 10
7. A golf ball of mass 0.05 kg placed on a tee, is
3. A body of mass m is thrown vertically upwards struck by a golf club. The speed of the golf ball
with a velocity v. It reaches to a height h and as it leaves the tee is 100 m/s, the time of contact
returns after t seconds. Then, the total change in between them is 0.02s. If the force decreases to
its momentum is – zero linearly with time, then the force at the
(A) zero (B) mv beginning of the contact is –
(C) mgt (D) mght (A) 500 N (B) 250 N
(C) 200 N (D) 100 N
4. A machine gun fires bullets of 50gm at the speed
of 1000 m/sec. If an average force of 200N is 8. A force F = 6t 2ˆi + 4tˆj , is acting on a particle of
exerted on the gunner, the number of bullets fired mass 3 kg then what will be velocity of particle at
per minute is – t = 3 sec. if at t = 0, particle is at rest-
(A) 240 (B) 120 (A) 18î + 6ˆj (B) 18î + 12ˆj
(C) 60 (D) 30 (C) 12iˆ + 6ˆj (D) None
, 2
9. A cricket player catches a ball of mass 100g 13. A body of mass 2 kg moves with acceleration
moving with a speed of 25 ms–1. If the ball is 3 ms–2. The change in momentum in one second
caught in 0.1s, the average force of the blow is:
2 3
exerted on the hands of the player is (A) kg ms−1 (B) kg ms−1
3 2
(A) 4 N (B) 25 N
(C) 6 kg ms−1 (D) None of these
(C) 40 N (D) 250 N
14. In case of a box lying on a horizontal table
10. The linear momentum P of a particle varies with
(A) Action and reaction are equal and opposite
time as follows, P = + t 2 , where and are and act perpendicular to the surfaces in
constants. The net force acting on the particle is contact
(A) Proportional to t (B) Proportional to t2 (B) Action and reaction are equal but act in the
(C) Zero (D) Constant same direction
(C) Action and reaction are not equal but are in
opposite directions
11. A machine gun fires n bullets per second and the
(D) Action of box on table and reaction of table
mass of each bullet is m. If v is the speed of each
on box are equal and opposite and are
bullet, then the average force exerted on the
inclined to vertical
machine gun is
(A) mng (B) mnv
15. The displacement (x) – time (t) curve of a particle
(C) mnvg (D) (mnv)/g
is shown in figure. The external force acting on
the particle is:
12. What is the impulse of force shown in the
Displacement ( x)
following figure?
F(N)
100 time (t)
O
(A) Acting at the beginning part of motion
t(s)
1 10 (B) Zero
(A) 225Ns (B) 450Ns (C) Not zero
(C) 900Ns (D) 1000Ns (D) None of these
