Question 1 1 / -0
A boy sitting on the top most berth in the compartment of a train, which is just going to stop at a railway station, drops an apple aiming at the open hand of his brother sitting vertically below him at a distance of about 2 m. The apple will fall
Solution
When the boy drops the apple, its initial velocity in the horizontal direction is the same as that of the train and there is no force acting in horizontal direction, so its velocity remains the same, whereas the train's motion is retarded. Hence, it will slow down so that the distance travelled by the train is less as compared to the apple in horizontal direction. So, the apple will fall away from the hands of his brother in the direction of motion of the train.
Question 2 1 / -0
A lead ball strikes a wall and falls down. A tennis ball having the same mass and same velocity strikes the same wall and bounces back. Which of the following statements is correct?
Solution
Change in momentum is given by Δp = mv - mu, where 'v' is the final velocity and 'u' is the initial velocity.
Question 3 1 / -0
Three blocks of masses m
1 = m, m
2 = 2m and m
3 = 3m, connected by two strings, are placed on a horizontal frictionless surface, as shown in the figure. A horizontal force F is applied to mass m
1 as shown.
The force on mass m
3 is
Solution
If F
2 and F
3 are the forces on masses m
2 and m
3 respectively, then the free-body diagrams of m
1 , m
2 and m
3 are as shown in Fig, where a is the common acceleration of the system.
F - F
2 = m
1 a (i)
F
2 - F
3 = m
2 a (ii)
F
3 = m
3 a (iii)
Adding Eqs. (i), (ii) and (iii) we get
Adding Eqs. (ii) and (iii) we have
F
2 = (m
2 + m
3 )a = (2m + 3m) ×
From Eq. (iii), we have
F
3 = 3ma = 3m ×
Question 4 1 / -0
A parrot is in a cage, which is hanging from a spring balance. Initially, the parrot sits in the cage; and in the second instance, the parrot flies about inside the cage. The reading of the balance
Solution
Let 'm' be the mass of the parrot and 'M' be the mass of the cage.
When the parrot is sitting in the cage, then the reading of the spring balance is N.
N = Mg + mg
When the parrot flies, the reaction of the surface of the cage on the parrot becomes zero.
Hence, new reading, N' = mg < N
Question 5 1 / -0
Two unequal masses are tied together with a compressed spring. When the cord is burnt with a match, releasing the spring, the two masses fly apart with equal
Solution
Using conservation of linear momentum, Momentum before burning of cord = Momentum after burning of cord
Momentum before burning of cord = 0
Momentum after burning of cord = mv + m'u
Here, m and m' are the masses of two particles, and u and v are their respective velocities.
mv + m'u = 0
Hence, the magnitudes of linear momenta of both masses are the same.
Question 6 1 / -0
A monkey weighing 20 kg slides down a vertical rope with a constant acceleration of 7.0 ms-2 . If g = 10 N kg-1 , the tension in the rope will be
Solution
Force acting on the monkey, mg = 20 x
10 = 200 N downwards
Using Newton's second law,
ma = mg - T
T = mg - ma = 200 - (20 x 7) = 60 N
Question 7 1 / -0
A force-time graph for the motion of a body is shown in figure. Change in linear momentum between 0 and 8 s is
Solution
Area under the force-time graph gives the change in linear momentum. Area above the time axis is taken as positive and area below the time axis is taken as negative.
Question 8 1 / -0
Two balls, each of mass 0.25 kg, moving with 3 m/s and 1 m/s towards each other in a straight line, collide. The balls stick together after the collision. The magnitude of the final velocity of the combined mass is
Solution
Let v be the velocity of the combined mass.
Initially, both balls are moving towards each other. The direction of the velocity of one mass will be taken as negative.
Hence, initial momentum,
p
i = 0.25 x 3 + 0.25 x (-1) = 0.5 kg m/s
Final momentum,
p
f = (0.25 + 0.25)v = 0.5v
0.5v = 0.5
v = 1 m/s
Question 9 1 / -0
Statement 1: While walking on the ice, person takes small steps instead of running to avoid slipping.Statement 2: While walking on the ice in small steps, he increases the normal downward force, which increases the normal reaction, which in turn increases the friction.
Solution
As the person takes small steps while walking on ice, the normal reaction of the surface is more as he pushes the ice perpendicular to the surface of the ice. Hence, the friction increases and the tendency of slipping decreases.
Question 10 1 / -0
A fast bowler bowls at 144 km/hr at the batsman, the batsman moves forward and hits back the ball with a speed of 162 km/hr towards bowler. What will be the magnitude of force exerted by the bat on the ball if the time of the contact is 1 millisecond?
Solution
Speed of the ball when it strikes the bat,
Speed of the ball after it rebounds off the bat is v,
So,
Change in linear momentum (take final direction as +ve),
Force exerted by the surface on the ball
F = 13.6 kN
Question 11 1 / -0
Two masses of 10 kg and 20 kg are connected by a massless spring. When a force of 200 N is applied on a 20 kg mass, the 10 kg mass accelerates at 12 m/sec
2 . What is the acceleration of the 20 kg mass?
Solution
F = m1 a1 + m2 a2 2 2 2 2 = 4 m/s2
Question 12 1 / -0
From a tower of height 20 m, a boy throws a stone in the vertically upward direction with a velocity of 40 m/s and at the same time a girl drops another identical stone from the same tower. When the momentum of the stone dropped by the girl is maximum, what will be displacement of the stone projected in the upward direction from the top of the tower?2 )
Solution
The momentum of the stone dropped will be maximum when it reaches the ground. Time taken by the stone dropped by the girl is t.
Using
(As u = 0 and downward direction is taken as positive)
t = 2 s
Displacement of the stone thrown vertically upwards,
(Taking upward direction as positive)
Question 13 1 / -0
A large truck and a car moving with same velocity have a head on collision. Which of the following is an INCORRECT statement?
Solution
According to Newton's third law, 'Every action has equal and opposite reaction'. So, both the car and the truck will experience the equal force of impact.
Question 14 1 / -0
A block is placed on a rough horizontal surface. A time dependent horizontal force F = kt acts on the block, where k is the positive constant. Acceleration-time graph for the block is
Solution
An object will not move until the force acting on the given body is less than the limiting friction. Hence, as long as the external force is less than the limiting friction, the acceleration of the object will be zero. When the applied force is greater than the limiting friction, constant kinetic friction will act. So, the acceleration of the block will be given by:
Question 15 1 / -0
A particle of mass m is tied to a light string and rotated with a speed v along the circular path of the radius 'r' in horizontal plane. If T is the tension in the string and mg = gravitational force on the particle, then pick the correct option(s).
Solution
Weight mg is the gravitational force with which the object is pulled towards the centre of the earth. Tension in the string is acting towards the centre of the circle which provides the necessary centripetal force.
Question 16 1 / -0
Two blocks initially at rest, having masses m1 = 1 kg and m2 = 2 kg moving under the action of external forces, attain the velocities of 5 m/s and 10 m/s in 5 s. Ratio of the forces acting on the two bodies is
Solution
Let the acceleration of block-1 be a
1 and the acceleration of block-2 be a
2. (as v = u + at)
Question 17 1 / -0
Which of the following assumptions is/are made in the law of conservation of momentum?
Solution
According to the law of conservation of momentum, in an isolated system (no external unbalanced forces are acting), the total momentum remains conserved.
Question 18 1 / -0
An object of mass 2 kg travelling in a straight line with a velocity of 9 ms-1 collides with and sticks to a stationary wooden block of mass 4 kg. What is the velocity of both the combined objects if they both move together in the same straight line after collision?
Solution
According to the law of conservation of linear momentum:
m
1 u
1 + m
2 u
2 = m
1 v
1 + m
2 v
2 (2 × 9) kg-ms
- 1 + (4 × 0) kg-ms
- 1 = (6 kg)v
v =
= 3 ms
-1
Question 19 1 / -0
A ball of mass 0.3 kg coming towards a player with a velocity of 10 ms
-1 is kicked back with a velocity of 20 ms
-1 in opposite direction. If the impact lasts for
seconds, the average force involved is
Solution
m = 0.3 kg, u = 10 m/s, v = -20 m/s
Impulse = change in momentum
F
t = mv - mu
F =
F = 0.3
Negative sign indicates that the direction of the force exerted on the ball is opposite to the direction of the initial velocity.
Magnitude of the force applied = 270 N
Question 20 1 / -0
A car moving at a speed 'v' is stopped by a retarding force F in distance 's'. If the speed of the car were 3v, then the force needed to stop it within the same distance 's' would be
Solution
Let m be the mass of the car and a be the deceleration produced by force F, then F = ma, where a is given by
2as = v
2 or a =
Therefore, F =
. Thus F ∝ v
2 .
If v becomes 3 times, F will become 9 times.
Hence, the correct choice is (3).
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