Motion In A Straight Line – NEET UG PhysicsPractice Questions & PYQs

When an object moves with uniform acceleration, the kinematic equation $v^2 = u^2 + 2as$ provides a direct link between velocity, initial conditions, and displacement. This principle dictates that the square of the velocity at any point is proportional to the distance traveled from the starting point, allowing us to determine intermediate velocities by analyzing specific segments of the motion.

For the entire path of length $s$ between points $P$ and $Q$, substituting the velocities 30 km/h and 40 km/h into the equation gives $40^2 = 30^2 + 2as$, which simplifies to $1600 = 900 + 2as$, meaning $2as = 700$. Focusing on the segment from $P$ to the midpoint $L$, where the displacement is $s/2$, we apply the same relation as $v^2 = 30^2 + 2a(s/2)$. By substituting the previously found value of $as = 350$ into this expression, we obtain $v^2 = 900 + 350 = 1250$, which results in a velocity of $25 \sqrt{2}$ km/h at the midpoint.

The velocity of a falling object just before impact is determined by the kinematic formula $v = \sqrt{2gh}$. Given a drop height of 5 m and gravitational acceleration of 10 m/s$^2$, the impact velocity is 10 m/s.

After the bounce, the rebound velocity $u$ is calculated using the same principle based on the height of 1.8 m, resulting in a velocity of 6 m/s. The velocity lost during the impact is the difference between the impact velocity and the rebound velocity, which amounts to 4 m/s. The fractional loss is found by comparing this decrease to the initial impact velocity, yielding $\frac{4}{10}$, which simplifies to $\frac{2}{5}$.

The motion of an object falling under gravity is governed by the kinematic relation $v^2 = u^2 + 2gh$. For the first object starting from rest, the initial velocity is zero, meaning the expression simplifies to $v^2 = 2gh$. Given that the final velocity is 3 m/s, the value of $2gh$ is 9.

For the second object, which starts with an initial velocity of 4 m/s and covers the same vertical displacement, we use the same kinematic relation $v^2 = u^2 + 2gh$. Substituting our known values gives $v^2 = 4^2 + 9$, which simplifies to $v^2 = 25$. Therefore, the final impact velocity of the second mass is 5 m/s.