In the given figure, the position-time graph of a particle of mass \$0.1\;\text{kg}\$ is shown. The impulse at \$\text{t}=2\;\text{sec}\$ is:

\$\textbf{Step 1: Initial velocity }\$Velocity is given by the slope of the displacement-time graph.From \$t=0 \ to \ t =2s\$ Slope =\$\dfrac{\text{Displacement}}{\text{time}}= \dfrac{4-0}{2-0}= 2\$ \$\therefore u = 2m/s\$ \$\textbf{Step 2: Final velocity}\$ At \$t = 2s\$, Slope of \$x - t\$ graph is zero \$\therefore v= 0m/s\$\$\textbf{Step 3: Impulse}\$Impulse is defined as change in momentum. \$I = m(v-u)\$ \$= 0.1 kg\times (0-2m/s)\$ \$= -0.2kg \ m \ s^{-1}\$Hence, Option(B) is correct.

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Newton's Second Law

Q: A machine gun fires a bullet of mass \$40g\$ with a velocity \$1200m/s\$. The man holding it can exert a maximum force of \$144N\$ on the gun. How many bullets can be fired per second at the most?

\$\textbf{Step 1: Newton's Second Law,}\$From Newton's second law \$F = \dfrac{dP}{dt}\$ \$....(1)\$Where, \$P=mv\$ is the momentum \$\textbf{Step 2: Total bullets fired per sec}\$Let n be the total number of bullets firedThe force exerted by man must balance the change in momentum of the bullets per second. \$\therefore P = nmv\$ \$;\$ \$m\$ and \$v\$ are constant (Given) \$\Rightarrow F = \dfrac{d}{dt} (nmv)\$ \$\Rightarrow\$ \$\dfrac{dn}{dt} = \dfrac{F}{mv} = \dfrac{144 (N)}{40 \times 10^{-3}(kg) \times 1200(m/s)} = 3\$Hence total number of bullets fired per second i.e., \$\dfrac{dn}{dt} = 3\$.Hence, Option \$B\$ is correct

Q: State and prove the law of conservation of momentum.

Law of conservation of momentum states that total momentum of system remains conserved in the absence of external force. Proof: Consider a body of mass m1 moving with velocity U1, striking against another body of mass m2 moving with velocity U2. Let the two bodies remain in contact with each other for a small interval "delta t". Let F12 be the average force exerted by mass m1 on m2, and let F21 be the force on m2 due to m1. Let v1 and v2 be the velcoities of two bodies after collision.Momentum of mass m1 before collision=m1u1Momentum of mass m2 after collision=m2u2Momentum of mass m1 after collision=m1v1By using the definition of impulse, change in momentum of mass m1 is,(refer figure).

Q: A \$0.5kg\$ ball moving with a speed of \$12m/s\$ strikes a hard wall at an angle of \${30}^{o}\$ with the wall. It is reflected with the same speed and at the same angle, as shown in figure. If the ball is in contact with the wall for \$0.25s\$ the average force acting on the wall is

\$F_{avg}=m(v_f-v_i)/t\$\$v_f=0m/s\$\$v_i=12m/s\$Therefore,\$F_{avg}=m(v_f-v_i)/t\$ \$=0.5(0-12)/0.25\$ \$=-24N\$

Q: A cricket ball of mass 150 g is moving with a velocity of 12 m/s and is hit by a bat so that the ball is turned back with a velocity of 20 m/s. The force of the blow acts for 0.001 & on the ball. Find that force?

\$Impulse = F\times t = \$ Change in Momentum\$\Rightarrow \$ Change in Momentum \$= mv-mu = 0.15(12)- 0.15(-20)= 4.8\$\$\Rightarrow F\times 0.001 = 4.8\$ \$\Rightarrow F = 4800N \$ Therefore, A is correct option.

Q: A rigid ball of mass m strikes a rigid wall at \$60^o\$ and gets reflected without loss of speed as shown in the figure below. The value of impulse imparted by the wall on the ball will be :

Initial momentum of the ball perpendicular to the wall \$P_i = m(-V cos60^o) = \dfrac{-mV}{2}\$Final momentum of the ball perpendicular to the wall \$P_f = m(V cos60^o) = \dfrac{mV}{2}\$Thus impulse imparted to the wall \$I = P_f - P_i = \dfrac{mV}{2} - \dfrac{-mV}{2} = mV\$

Q: A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of \$12 ms^{-1}\$. If the mass of the ball is \$0.15\ kg\$, determine the change in momentum of the ball. (Assume linear motion of the ball)

The initial momentum of ball \$P_1 = m\vec{v} \$, after batsman hits ball and reverses its direction, momentum of ball \$P_2= -m\vec{v} \$Magnitude of change in momentum \$\Delta p = \left | P_2-P_1\right | =\left | - 2m \vec{v}\right | =2m \left | \vec{v}\right | \$ from data given in question \$ m=0.15 kg ;\; \left | \vec{ v} \right |= 12m/ s\$\$\Delta P=2\times 0.15\times 12= 3. 6 kg m/ s \$

Q: A hockey ball of mass 200 g travelling at \$10\ ms^{-1}\$ is struck by a hockey stick so as to return it along its original path with a velocity of \$5\ ms^{-1}\$. Calculate the change of momentum which occurred in the motion of the hockey ball by the force applied by the hockey stick:

Mass of ball(m)\$=\$200g\$=\$0.2kgInitial velocity of ball(u)\$=\$10m\${s}^{-1}\$Final velocity of ball(v)\$=\$5m\${s}^{-1}\$Therefore, change in momentum (\$\triangle\$p) \$=\$ Final momentum - initial momentum \$=m(v-u)=0.2\times(-5-10)=-3\ kgm{s}^{-1}=-3\ Ns\$

Q: A force time graph for the motion of a body is shown in figure. Change in linear momentum between \$0\$ and \$8s\$ is :

The area under the graph gives linear momentumLinear momentum=Force×timeChange in momentum=\$(-2 \times 2)+(4\times 1)+(-1\times 1)+(1\times 1)=5-5=0\$Change in momentum=0

Q: If bullet of mass \$5\$ \$gm\$ moving with velocity \$100\$ \${m/sec}\$, penetrates the wooden block upto \$6\$ \$cm\$. Then the average force imposed by the bullet on the block is

\$F=ma\$\$a=\cfrac{v}{t}\$ and \$t=\cfrac{x}{v}\$Mass in \$kg\$,\$a=\cfrac{v^2}{x}=\cfrac{100^2}{6\times 10^{-2}}\$\$F=\cfrac{5}{1000}\times \cfrac{100^2}{6\times 10^{-2}}=833.33\$ \$N\$

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