Numerical Problems Based on Linear Momentum and Newton’s Second Law of Motion for Class 11 Physics

After the force ceases, the body moves with a constant velocity. We can find this velocity ($v$) from the distance it covers in the next $5\text{ s}$:

This means the force accelerated the body from rest ($u = 0$) to a velocity of $10\text{ ms}^{-1}$ in $t = 10\text{ s}$.

Calculate the acceleration ($a$):

Now, calculate the magnitude of the force ($F$):

Initial velocity $u = 0$, distance $s = 400\text{ m}$, time $t = 20\text{ s}$.
Mass $m = 7\text{ metric tonnes} = 7 \times 1000 = 7000\text{ kg}$.

First, calculate the acceleration ($a$) using the second equation of motion:

Next, calculate the force ($F$):

Given Force $F = 5\text{ N}$.

Calculate individual masses using $m = \frac{F}{a}$:

When the masses are tied together, the total mass is $M = m_1 + m_2$:

The new acceleration ($a_{new}$) with the same force is:

Force $F = 10^{-23}\text{ N}$, Mass $m = 10^{-30}\text{ kg}$, Distance $s = 0.1\text{ m}$. Assume initial velocity $u = 0$.

First, find the acceleration ($a$) of the electron:

Now, calculate the time ($t$) taken to cover the distance using $s = ut + \frac{1}{2}at^2$:

Mass $m = 5\text{ kg}$, Height $h = 50\text{ m}$, Penetration depth $s = 1\text{ m}$. Take $g = 9.8\text{ ms}^{-2}$.

1. Velocity just before hitting the sand ($v_1$):

2. Motion inside the sand:
Initial velocity $u’ = \sqrt{980}\text{ ms}^{-1}$, final velocity $v’ = 0$, distance $s = 1\text{ m}$.
Find the retardation ($a’$) inside the sand:

3. Resistance offered by the sand:
The net force causing this deceleration is the resistance minus gravity (though standard textbook approximations often equate net force directly to the resistance if gravity’s contribution during the 1m is negligible compared to the impact force).

4. Time to penetrate ($t$):

Mass $m = 100\text{ g} = 0.1\text{ kg}$, Initial velocity $u = 20\text{ ms}^{-1}$, Final velocity $v = 0$.
Penetration distance $s = 20\text{ cm} = 0.2\text{ m}$.

First, find the retardation ($a$) using the third equation of motion:

Now, calculate the resistance force ($F$) offered by the plank: