Table of contents
MCQ questions on force and laws of motion
Short answer type questions with answers
Numerical questions with answers
MCQ (Multiple choice questions) on Force and laws of Motion
1. Motion of rocket is based on the principle of
(a) conservation of mass, (b) conservation of energy, (c) conservation of force, (d) conservation of linear momentum
option d
The motion of rocket is based on the conservation of linear momentum.
2. When a body moves at constant velocity, which of the following is true
(a) A constant force is acting on the body to continue its motion
(b) Net force is zero
(c) Moves with a constant acceleration
(d) None of the above
option b
When a body moves at constant velocity, the acceleration of the body is zero, so net force = mass \times acceleration, which is also zero.
3. Rate of change of momentum is called
(a) Pressure, (b) Acceleration, (c) Force, (d) Work
option c
From Newtons second law,
force is defined as the rate of change of momentum.
4. A 2 kg ball rolls around the edge of a circle with a radius of 1m. If it is rolling at a speed of 3m/s, the centripetal acceleration is
(a) 9 𝑚𝑠-2, (b) 18 𝑚𝑠-2, (c) 6 𝑚𝑠-2, (d) None of these
option a
Mass of the body is, m = 2 kg
radius of the circular path, r = 1 m
speed, v = 3 m/s
Centripetal acceleration,
a=\frac{v^{2}}{r}
a=\frac{3^{2}}{1} = 9 𝑚𝑠-2
5. The unit of impulse is
(a) N, (b) N.s, (c) N.m, (d) None of the above
option b
Impulse = Force \times time
The unit of force is newton and that of time is second,
So, the unit of impulse is Ns (option b)
Short answer type questions with answers on force and laws of motion
1. State the law of conservation of momentum. Give one example of conservation of linear momentum.
The principle of conservation of linear momentum states that if the net force acting on a system of bodies is zero, then the total linear momentum of the system of bodies remain constant.
Some examples of conservation of linear momentum are the explosion of a bomb or the recoil velocity of a gun.
2. What is the difference between impulse and impulsive force?
3. Impulse: Impulse is the product of force and the time interval for which the force acts on the object.
If F is the applied force for a time interval t, then impulse,
I = F \times t
I = ma \times t [Force = mass (m) \times acceleration (a)]
I=m\frac{(v-u)}{t} \times t [u is the initial velocity and v is the velocity after time t]
I=mv-mu
I= change of momentum
Impulsive force: Impulsive force is a force which acts on a body for very short period of time.
Numerical problems with solution on Force and Motion
1. A body of mass 10 kg is moving with a velocity of 40m/s. Calculate the force required to bring the body at rest in 5 s.
Mass of the body, m = 10kg
Initial velocity, u = 40 m/s
Final velocity, v = 0 m/s
time, t = 5 s
Force, F = mass (m) \times deceleration (a)
F= m \times \frac{(u-v)}{t}
F= 10 \times \frac{(40-0)}{5}
F = 80 N
2. A force acts for 10 s on a body of mass 5 kg which was initially at rest. The body moves uniformly and travels a distance of 15m in the next 5 s. Calculate the magnitude of the force.
mass, m = 5 kg, initial velocity, u = 0 m/s
The body moves a distance, s = 15 m in 5 s at a uniform (constant) velocity, so
final velocity, v = s/t = 15/5 = 3 m/s
F = mass (m) \times acceleration (a)
F= m \times \frac{(v-u)}{t}
The force acts for a time, t = 10 s
F= 5 \times \frac{(3-0)}{10}
F= 15/10 N = 1.5 N
3. A 10 kg gun fires a bullet of mass 50 gm with a velocity of 400 m/s. Find the recoil velocity of the gun.
Mass of the gun, M = 10 kg
mass of the bullet, m = 50 gm = 50/1000 kg = 0.05 kg
velocity of the bullet, v = 400 m/s
Let the velocity of the gun is V.
From the conservation of linear momentum, the net momentum of the system (gun+bullet) before and after firing must be equal.
The momentum before the firing is zero. So, after the bullet is fired the total momentum should be zero.
After firing of the bullet the momentum of the system is MV+mv.
So, MV+mv=0
MV=-mv
V=-\frac{mv}{M}
V=-\frac{0.05 \times 400}{10}
V= -2 m/s
So, the recoil velocity of the gun is 2 m/s. [The negative sign shows that the bullet and the gun moves in opposite direction]
4. A bullet of 5 gm is fired from a pistol of 1.5 kg. If the recoil velocity of pistol is 1.5 m/s, find the velocity of bullet.
Mass of the pistol, M = 1.5 kg
mass of the bullet, m = 5 gm = 5/1000 kg = 0.005 kg
recoil velocity of the pistol, V =1.5 m/s
velocity of the bullet, v = ?
From the conservation of linear momentum, the net momentum of the system (pistol+bullet) before and after firing must be equal.
The momentum before the firing is zero. So, after the bullet is fired the total momentum should be zero.
After firing of the bullet the momentum of the system is MV+mv.
So, MV+mv=0
mv = -MV
v=-\frac{MV}{m}
v=-\frac{1.5 \times 1.5}{0.005}
v= -450 m/s
So, the velocity of the bullet is 450 m/s. [The negative sign shows that the bullet and pistol moves in opposite direction]
5. To launch a 6000kg rocket with an upward acceleration of 20m/s2, how much kg/s of fuel has to be burnt out to eject burnt out gas with a velocity 900 m/s? take g= 10m/s2
The mass of the rocket, M=6000 kg
acceleration, a = 20m/s2
acceleration due to gravity, g = 10m/s2
The speed of the burnt out fuel, v = 900 m/s
The upthrust, F=v\frac{dm}{dt}
Again, F = M(a+g) = 6000 \times (20+10) = 180000 N
So, v\frac{dm}{dt} = 180000
\frac{dm}{dt}=\frac{180000}{v}
\frac{dm}{dt}=\frac{180000}{900}=200 kg/s
Therefore, 200 kg/s of fuel has to be burnt out to get an upward acceleration of 20m/s2 .