Practice free Laws of Motion (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
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 $
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