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1. Which of the following has more inertia:
(a) A rubber ball and a stone of the same size?
(b) A bicycle and a train?
(c) A five rupees coin and a one-rupee coin?
Ans: (a) A stone has more inertia
(b) A train has more inertia
(c) A Five rupees coin has more inertia
2. In the following example, try to identify the number of times the velocity of the ball changes:
“A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team”.
Also identify the agent supplying the force in each case.
Ans: In the above case the velocity of football changed for 4 times.
|
Action |
Agent |
|
The football player kick the ball for first time |
Legs |
|
The football was kicked by another player |
Legs |
|
The goalkeeper catches the football |
Hands |
|
The goalkeeper kicks the football |
Legs |
3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Ans: When the tree is at rest its leaves along with branches remains at rest but when we vigorously shake its branches, the branches came to motion but the inertia in leaves remains at rest due to which the leaves tries to stay at its original position and they gets detached from the branches.
4. Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?
Ans: When the bus is moving, our inertia is set to motion and hence when the bus stops we fall in forward direction. Similarly, when bus is at rest our inertia is in rest, thus when the bus starts we fall backwards.
5. If action is always equal to the reaction, explain how a horse can pull a cart.
Ans: When the horse asserts force on the ground, the ground in turn push the horse forward, this is how it is able to pull the cart.
6. Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.
Ans: Since, every action has an equal and opposite reaction thus when the fireman hold the hose, which ejects large amount of water. The hose exert greater force in the backward direction, which makes it difficult to hold the hose.
7. From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s–1 . Calculate the initial recoil velocity of the rifle.
Ans: The mass of rifle m1 = 4kg
The mass of bullet m2 = 50g
The initial velocity of rifle u1 = 0 m/s
The initial velocity of rifle u2 = 0 m/s
Let the final velocity of the rifle be v1
The final velocity of the bullet is v2 = 35 m/s
According to law of conservation of momentum
m1u1 + m2u2 = m1v1 + m2v2
=> 4 x 0 + (50/100) x 0 = 4v + (50/1000) x 35
=> 0 = 4v + 35/20
=> - 35/20 = 4v
=> -35/80 = v
Therefore, the initial recoil velocity of rifle is -35/80 m s-1
8. Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 m s–1 and 1 m s–1, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 m s–1 . Determine the velocity of the second object.
Ans: The mass and initial velocity of first object is m1 = 100g and u1 = 2 m/s respectively.
The mass and initial velocity of second object is m2 = 200g and u2 = 1 m/s
The final velocity of first object v2 = 1.67 m/s
According to law of conservation of momentum
m1u1 + m2u2 = m1v1 + m2v2
=> 100 x 2 + 200 x 1 = 100 x 1.67 + 200 x v2
=> 200 + 200 = 167 + 200 v2
=> 400 = 167 + 200 v2
=> 400 – 167 = 200 v2
=> 233 = 200 v2
=> 233/200 = v2
=> 1.165 m/s = v2
Therefore, the velocity of second object after collision is 1.165 m/s.
1. An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.
Ans: Yes, it is possible for an object to travel with a non-zero velocity if:
2. When a carpet is beaten with a stick, dust comes out of it. Explain.
Ans: When the carpet is beaten, it comes in motion, whereas due to inertia the dust remains at rest and we’re able to see it.
3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Ans: When the bus is in motion, the luggage kept on the roof of it remains at rest due to inertia. So, the if the moving bus stops the luggage may fall down. Thus, it is advised to tie it with a rope at the roof.
4. A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because
(a) the batsman did not hit the ball hard enough.
(b) velocity is proportional to the force exerted on the ball.
(c) there is a force on the ball opposing the motion.
(d) there is no unbalanced force on the ball, so the ball would want to come to rest.
Ans: (c) there is a force on the ball opposing the motion.
5. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes (Hint: 1 tonne = 1000 kg.)
Ans: The initial velocity of the truck u = 0 m/s
The truck travels 400m in 20s.
Therefore, it’s final velocity would be v = 400/20
= 20 m/s
Now, s = ut + ½ at2
=> 400 = 0 x 20 + ½ a x (20)2
=> 400 = 0 + ½ a x 400
=> 400 = 200a
=> 400/200 = a
=> 2 = a
So, acceleration is 2 m/s.
Now, if the mass of truck is 7 tonne or 7000 kg.
Therefore, the force acting on it would be F = m x a
=> F = 7000 x 2
=> F = 14000 N
So, the force exerted on the truck is 14000 N.
6. A stone of 1 kg is thrown with a velocity of 20 m s–1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Ans: Given the initial and final velocity of the stone is u = 20 m/s and v = 0m/s respectively.
Distance covered by the stone s = 50m
Now, using 2as = v2 – u2
=> 2a(50) = (0)2 – (20)2
=> 100a = - 400
=> a = -400/100
=> a = -4 m/s
So, the acceleration of the stone is -4 m/s2 (retardation).
Now, the Force F = m x a
=> F = 1 x -4
=> F = -4N
Therefore, the force exerted was -4N.
7. A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate:
(a) the net accelerating force and
(b) the acceleration of the train.
Ans: Given, the force exerted by the engine is 40000N and the friction force offered by the track is 5000N.
(a) So, the net accelerating force F = 40000N – 5000N
F = 35000 N
(b) The total mass of the train m = mass of engine + mass of wagon x 5
m = 8000 + (2000 x 5)
m = 8000 + 10000
m = 18000kg
Now, the acceleration of the train a = F/m
=> a = 35000/18000
=> a = 1.944 m/s2
8. An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 m s–2?
Ans: Given the mass of vehicle is m = 1500kg and the negative acceleration a = 1.7 m s-2
So, Force F = ma
=> F = 1500 x (-1.7)
=> F = - 2550 N
So, the force between the vehicle and road is -2550N.
9. What is the momentum of an object of mass m, moving with a velocity v?
(a) (mv)2 (b) mv2 (c) ½ mv2 (d) mv
Ans: (d) mv
10. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Ans: Since, the velocity of the cabinet is 0. Therefore, its acceleration of will also be zero. The frictional force acts in the opposite direction of motion. Hence, the frictional force exerted on the cabinet would be -200N.
11. Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 m s-1 before the collision during which they stick together. What will be the velocity of the combined object after collision?
Ans: Given, the mass of one object m1 = 1.5kg
And the mass of second object m2 = 1.5kg
The velocity of first object is u1 = 2.5 m/s
The velocity of second object is u2 = -2.5 m/s
Using law of conservation of momentum = m1u1 + m2u2 = m1v1 + m2v2
=> (1.5 x 2.5) + (1.5 x 2.5) = (1.5 x v) + (1.5 x v)
=> 3.75 – 3.75 = (1.5 + 1.5) v
=> 0 = 3v
=> v = 0
Therefore, the combined velocity of the objects after collision is 0 m/s.
12. According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
Ans: The logic given by the student is incorrect because action and reactions acts on different bodies so it will not cancel. The truck parked along the roadside has heavy weight which means high inertia. Due to which the truck will not move. To make the truck move greater force will be required.
13. A hockey ball of mass 200 g travelling at 10 m s–1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 m s–1. Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Ans: Given, the mass of the hockey ball is 200g and its velocity initial velocity is 10 m/s.
So, the momentum of the ball would be p1 = mu
=> p1 = 200 x 10
=> p1 = 2000 g m/s
Now, after hit by the hockey stick, the velocity of the ball become v = -5 m/s
So, the momentum after hit by the hockey would be p2= mv
=> p2 = 200 x 5
=> p2 = 1000 g m/s
Now, change in the momentum would be p2 – p1
= -1000 – 2000
= -3000 g m/s
14. A bullet of mass 10 g travelling horizontally with a velocity of 150 m s–1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
Ans: Given, mass of bullet m = 10g = 0.01kg
Initial velocity u = 150 m/s
Final velocity v = 0
Time t = 0.03s
Now, using v = u + at
=> 0 = 150 + (a x 0.03)
=> -150 = 0.03a
=> -150/0.03 = a
=> -15000/3 = a
=> -5000 m/s2 = a
Now, using v2 = u2 + 2as
=> (0)2 = (150)2 + 2(-5000)xs
=> 0 = 22500 – (10000 x s)
=> -22500 = -10000 x s
=> -22500/ -10000 = s
=> 2.25 m = s
So, the distance travelled by bullet is 2.25 m.
Now, to calculate the Force F = ma
=>F = 0.01 x 5000
=> F = 50 N
So, the force exerted by the wooden block is 50 N.
15. An object of mass 1 kg travelling in a straight line with a velocity of 10 m s–1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
Ans: Given the mass of object is m1 = 1kg
Mass of wooden block m2 = 5kg
Initial velocity of object u1 = 10m/s
Initial velocity of wooden u2 = 0 m/s
So, total momentum just before collision m1u1 + m2u2
= (1 x 10) + (5 x 0)
= 10 + 0
= 10 kg m/s
Hence, the total momentum before collision is 10 kg m/s.
Now, from conservation of momentum
=> m1u1 + m2u2 = m1v + m2v
=> (1 x 10) + (5 x 0) = (m1 + m2) v
=> 10 + 0 = (1 + 5)v
=> 10 = 6v
=> 10/6 = v
=> 5/3 = v
Hence, the combined velocity is 5/3 m/s.
16. An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s–1 to 8 m s–1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Ans: Given, the mass of the object m = 100kg
Initial velocity u = 5 m/s
Final velocity v = 8 m/s
Time t = 6s
Now,
Initial momentum = mu = 100 x 5 = 500 kg m/s
Final momentum = mv = 100 x 8 = 800 kg m/s
Now, the force exerted on the object F = change in momentum / time taken
=> F = (800 – 500) / 6
=> F = 300/6
=> F = 50 N
18. How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 m s–2 .
Ans: Given, mass of dumb-bell m = 10kg
Initial velocity u = 0 m/s
Acceleration a = 10 m/s2
And distance travelled s = 80 cm = 0.8 m
Now, using v2 = u2 + 2as
=> v2 = (0)2 + 2(10)(0.8)
=> v2 = 0 + 16
=> v2 = 16
=> v = 4 m/s
Now, the momentum transfer by the dumb-bell to the floor
=> p = mv
=> p = 10 x4
=> p = 40 kg m/s