NEET Physics Questions: Laws of Motion
Pratcice NEET questions from all chapters from huge question bank for free. All MCQs are based on NCERT syllabus. To practice from a specific subject and chapter, select a subject below. Login to practice in a structured way with explanations, bookmarks, lists, notes etc. Click here to Login or Sign up for free.
Question bank:
Two pulley arrangements of figure given are identical. The mass of the rope is negligible. In fig (a), the mass m is lifted by attaching a mass 2m to the other end of the rope. In fig (b), m is lifted up by pulling the other end of the rope with a constant downward force F = 2mg. The acceleration of m in the two cases are respectively
The velocity of a body of mass 20 kg decreases from $20 ms^{–1} to 5 ms^{–1}$ in a distance of 100 m. Force on the body is
$ F = ma = m \left( \nu^2 - u^2 \over 2S \right) $
A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes upto 2 m height further, find the magnitude of the force.$ (Consider g = 10 ms^{–2}) $
$ H_{max} = { u^2 \over 2g } $
$ \therefore u = \sqrt { 2gH_{max} }$
This velocity is supplied to the ball by hand and initially the hand was at rest. It acquires this velocity in distance of 0.2 meter.
$ \therefore a = { u^2 \over 2S } $
So upword force F = m (g+a)
Formula for true force is
According to newton's second law force = rate of change of linear momentum.
A particle moves in the X–Y plane under the influence of a force such that its linear momentum is $ \vec P (t) = A[ \hat i cos (kt) -\hat j sin(kt)] $ where A and k are constants .The angle betweenn the force and momentum is
$ \vec P (t) = A[ \hat i cos (kt) -\hat j sin(kt)] $
$ \vec F = { d \over dt} \left ( \vec P (t) \right) $
$ = Ak \left( - \hat i sin kt - \hat j cos ky \right) $
$ \vec F . \vec P = A^2 k ( -cos kt .sin kt + sin kt . cos kt ) $
$ \vec F . \vec P = 0 $
Force of 5 N acts on a body of weight 9.8 N. what is the acceleration produced in $ms^{-2}$
As weight W = mg = 9.8 N
Therefore m = 1 Kg
a = F / m
Same force acts on two bodies of different masses 2 kg and 4 kg initially at rest. The ratio of times required to acquire same final velocity is
$ t = {V \over a } \Rightarrow t \alpha { 1 \over a } ( v is constant ) $
$ \therefore {t_1 \over t_2 } = {a_2 \over a_1} = { m_1 \over m_2 } = {2 \over 4 } = { 1 \over 2 } $
$ a \alpha 1/m as F is constant $
Which of the following quantities measured from different inertial reference frames are same
Force is a quantity that is invariant under Galilean transformations between different inertial reference frames. This means that when observed from different inertial frames, the force remains the same. In contrast, quantities like velocity, displacement, and kinetic energy can vary depending on the reference frame.
When the speed of a moving body is doubled
A particle moves in the XY Plane under the action of a force F such that the components of its linear momentum P at any time t are Px = 2 cost, Py = 2 sint. The angle between F and P at time t is
$ \vec P = Px \hat i + Py \hat j , \vec F = { \vec {dp} \over dt} $
Now $ \vec F . \vec P = 0 $
$ \therefore \theta = 90 ^\circ $
A player caught a cricket ball of mass 150 g moving at the rate of $20 ms^{-1} $ . If the catching process be completed in 0.1 s the force of the blow exerted by the ball on the hands of player is
Force exerted by ball, F = m (dv /dt)
A body of mass 5 kg starts from the origin with an initial velocity $ \vec u = 30 \hat i + 40 \hat j ms^{-1} $. If a constant Force $ \vec F = - \left( \hat i + 5 \hat j \right) N $ acts on the body, the time in which the y-component of the velocity becomes
$ \nu_y = 40ms^{-1} Fy = -5N m = 5kg $
$ so a_y = { F_y \over m } = -1 ms^{-1} $
$ as V_y = u_y + at $
$ 0 = 40 -lt \Rightarrow t = 40 s $
A cold soft drink is kept on the balance.When the cap is open, then the weight
Gas will cane out with sufficient speed in forward direction, So recetion of this forward force wil change the reading of the spring balance.
A wagon weighing 1000 kg is moving with a velocity $50 km h^{-1}$ on smooth horizontal rails. A mass of 250 kg is dropped into it. The velocity with which it moves now is
According to the principle of conservation of linear momentum
$ 1000 \times 50 = {1250 \times \nu \over \nu} = 40 km/h $
A 7 kg object is subjected to two forces (in newton) $ \vec F _1 = 20 \hat i + 30 \hat j $ and $ \vec F_2 = 8 \hat i -5 \hat j $. The magnitutde of resulting acceleration in $ ms^{-2} $ will be
$ \vec F = \vec F_1 + \vec F_2 $
$ \therefore a = F /m $