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What is Momentum?

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: The force 'F' acting on a particle of mass 'm' is indicated by the force-time graph shown below. The change in momentum of the particle over the time interval from zero to 8 s is

\$\Delta P=\left (\dfrac {1}{2}\right )(2)(6)-(3)(2)+(4)(3)\$\$=6-6+12\$\$=12\$

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 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: 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: If a body of mass M collides against a wall with velocity V and rebounds with the same speed, then its change of momentum will be

Initial velocity of body before collision \$u = -V\$where minus sign indicates the body moves towards the wall.Initial momentum of the body \$P_i = Mu = -MV\$Final velocity of body after collision \$v =V\$where positive sign indicates the body moves away from the wall.Final momentum of the body \$P_f = Mv = MV\$Change in momentum \$\Delta P = P_f - P_i = MV - (-MV) = 2MV\$

Q: Two masses of 1 gm and 4 gm are moving with equal kinetic energies. The ratio of the magnitudes of their linear momenta is:

Kinetic energy \$K = \dfrac{p^2}{2m}\$Two masses are given as- \$m_1 = 1 \ gm\$ and \$m_2 = 4 \ gm\$But \$K_1 = K_2\$\$\therefore\$ \$ \dfrac{p^2_1}{2m_1} = \dfrac{P^2_2}{2m_2}\$Or \$ \dfrac{P_1^2}{2(1)} = \dfrac{P^2_2}{2(4)}\$\$\implies\$ \$P_1:P_2 = 1:2\$

Q: The moment P (in \$kg\, ms^{-1}\$) of a particle is varying with time t (in second) as \$p=2+3t^2\$. The force acting on the particle at \$t=3s\$ will be

Given,\$p=2+3t^2\$From Newton's second law,\$F=\dfrac{dp}{dt}\$\$F=\dfrac{d(2+3t^2)}{dt}\$\$F=0+6t=6t\$At \$t=3s\$\$F=6\times 3=18N\$The correct option is A.

Q: A force of 100 dynes acts on mass of 5 gm for 10 sec. The velocity produced is:

Given that,Force \$F=100\,dyn\$Mass \$m=5\,g\$Time \$t=10\,s\$Initial velocity \$u=0\$Now, the acceleration is \$ F=ma \$ \$ a=\dfrac{F}{m} \$ \$ a=\dfrac{100\times {{10}^{-5}}}{5\times {{10}^{-3}}} \$ \$ a=0.20\,m/{{s}^{2}} \$Now, from equation of motion \$ v=u+at \$ \$ v=0+0.20\times 10 \$ \$ v=2\,m/s \$ \$ v=200\,cm/s \$Hence, the velocity produced is \$200\ cm/s\$

Q: What is the momentum of an object of mass \$m\$, moving with a velocity \$v\$?

Hint:The momentum is defined as the product of the mass and the velocity.\$\textbf{Step 1: Showing the expression of momentum}\$The expression for the momentum is written as,\$ { p } =mv\$Here,\$ { p }\$ is the momentum, \${m}\$ is the mass and \${v}\$ is the velocity of the object.\$\textbf{Hence, option D is the correct answer.}\$

Physics

definition

Linear Momentum

Linear momentum is a vector quantity defined as the product of an object's mass,

m

, and its velocity,

v

. Linear momentum is denoted by the letter

p

and is called momentum for short. Note that a body's momentum is always in the same direction as its velocity vector. The unit of momentum are kg. m/s.

p = m v

definition

Unit and formula for momentum

Momentum is a product of mass and velocity.It is denoted as

p

. Momentum has both direction and magnitude. It is a vector quantity. Its direction is same as that of the velocity.

p = m v

Unit of momentum is

k g m / s

N s

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