Third Law of Motion and Momentum and Impulse Questions in English

(C) The momentum $p$ of a body of mass $m$ moving with velocity $v$ is defined as $p = mv$.
Since the mass $m$ of a body is a constant scalar quantity,if the momentum $p$ is constant,then the velocity $v$ must also be constant.
Therefore,for a body with constant momentum,its velocity remains constant.

(C) Newton's third law of motion states that for every action,there is an equal and opposite reaction. This law describes force as a mutual interaction between two bodies,where one body exerts a force on another,and the second body simultaneously exerts an equal and opposite force on the first.

(B) The impulse imparted to the ball is equal to the area under the $F-t$ graph.
Impulse = Area of the triangle = $\frac{1}{2} \times \text{base} \times \text{height}$
Impulse = $\frac{1}{2} \times (2 \times 10^{-3} \ s) \times (20 \times 10^3 \ N)$
Impulse = $20 \ N \ s = 20 \ kg \ m \ s^{-1}$
According to the impulse-momentum theorem,impulse is equal to the change in momentum $(\Delta p)$:
$\Delta p = m v - m u = \text{Impulse}$
Given: mass $m = 100 \ g = 0.1 \ kg$,initial velocity $u = 10 \ m \ s^{-1}$.
$0.1 \times v - 0.1 \times 10 = 20$
$0.1 \times v - 1 = 20$
$0.1 \times v = 21$
$v = \frac{21}{0.1} = 210 \ m \ s^{-1}$

(C) Given:
Momentum of the bullet $(p)$ $= 20 \,kg \cdot m/s$
Velocity of the bullet $(v)$ $= 1000 \,m/s$
We know that the formula for momentum is $p = m \times v$, where $m$ is the mass of the object.
Rearranging the formula to solve for mass: $m = \frac{p}{v}$.
Substituting the given values: $m = \frac{20}{1000} \,kg$.
$m = 0.02 \,kg$.
To convert the mass into grams, multiply by $1000$: $m = 0.02 \times 1000 \,g = 20 \,g$.
Therefore, the mass of the bullet is $20 \,g$.

(A) The average force $F$ exerted by the gun on the bullets is given by the rate of change of momentum: $F = \frac{\Delta p}{\Delta t} = n \times m \times v$,where $n$ is the number of bullets fired per second,$m$ is the mass of each bullet,and $v$ is the velocity of the bullet.
Given:
Number of bullets per minute $= 200$,so $n = \frac{200}{60} \ s^{-1}$.
Mass of each bullet $m = 35 \ g = 0.035 \ kg$.
Velocity $v = 750 \ m \ s^{-1}$.
Substituting the values:
$F = \left(\frac{200}{60}\right) \times 0.035 \times 750$
$F = \left(\frac{20}{6}\right) \times 35 \times 0.75$
$F = \frac{10}{3} \times 35 \times 0.75 = 10 \times 35 \times 0.25 = 87.5 \ N$.

(C) Given: Mass $m = 0.2 \ kg$,initial velocity $u = 20 \ m s^{-1}$,final velocity $v = 0 \ m s^{-1}$,and time $t = 0.1 \ s$.
Using the impulse-momentum theorem or Newton's second law,the average force $F$ is given by the rate of change of momentum:
$F = \frac{\Delta p}{\Delta t} = \frac{m(v - u)}{t}$
Substituting the values:
$F = \frac{0.2 \times (0 - 20)}{0.1}$
$F = \frac{0.2 \times (-20)}{0.1}$
$F = \frac{-4}{0.1} = -40 \ N$
The magnitude of the average force applied to the ball is $40 \ N$.

(D) The impulse imparted to the ball is equal to the change in its linear momentum.
Impulse $= \int F \, dt = \text{Area under the } F-t \text{ graph}$.
Given that the force decreases linearly from $F$ to $0$ in time $t = 0.02 \text{ s}$, the area is a triangle with base $0.02 \text{ s}$ and height $F$.
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 0.02 \times F = 0.01F$.
The change in momentum is $\Delta p = m(v_f - v_i) = \frac{50}{1000} \text{ kg} \times (100 \text{ m/s} - 0) = 0.05 \times 100 = 5 \text{ kg m/s}$.
Equating impulse and change in momentum: $0.01F = 5$.
$F = \frac{5}{0.01} = 500 \text{ N}$.

(D) Given: Mass $m = 0.25 \,kg$, initial velocity $u = 10 \,m/s$, final velocity $v = -10 \,m/s$ (since it returns in the opposite direction), and time interval $\Delta t = 0.01 \,s$.
Change in momentum $\Delta P = m(v - u) = 0.25 \times (-10 - 10) = 0.25 \times (-20) = -5 \,kg \cdot m/s$.
The magnitude of change in momentum is $|\Delta P| = 5 \,kg \cdot m/s$.
The force exerted on the ball by the bat is $F = \frac{\Delta P}{\Delta t} = \frac{5}{0.01} = 500 \,N$.
According to Newton's third law, the force exerted on the bat by the ball is equal in magnitude, which is $500 \,N$.