Centre of Mass, Momentum and Collision Concepts Page - 7

Example

Problem in Inelastic Collision

Example: Two bodies move towards each other and collide inelastically. The velocity of the first body before impact is 2$2$ m/s and of the second is 4$4$ m/s.The common velocity after collision is 1$1$ m/s in the direction of the first body. How many times did the K.E. of the first body exceed that of the second body before collision.

Solution:
using momentum conservation
m1×2−4m2=(m1+m2)$m_1\times 2-4m_2=(m_1+m_2)$
⇒m1=5m2$\Rightarrow m_1=5m_2$
Ratio of K.E=12×m1×(2)212×m2×(4)2$={\displaystyle \frac{{\displaystyle \frac{1}{2}}\times m_1\times (2{)}^2}{{\displaystyle \frac{1}{2}}\times m_2\times (4{)}^2}}$
=5m2×4m2×16$={\displaystyle \frac{5m_2\times 4}{m_2\times 16}}$
=1.25$=1.25$

Example

Inelastic Collision in two dimensions

Example: A particle of mass m$m$ is projected from the ground with an initial speed u0$u_0$at an angle α$\alpha$with the horizontal. A$A$t the highest point of its trajectory, it makes a completely inelastic collision with another identical particle, which was thrown vertically upward from the ground with the same initial speed u0$u_0$. What is the angle that the composite system makes with the horizontal immediately after the collision?

Solution:
Velocity of the particle performing projectile motion at highest point
=V1=V0cosα$=V_1=V_{0}cos\alpha$
Velocity particle thrown vertically upwards at the position of collision
=v22=u20−2gu20sin2α2g=u0cosα${\displaystyle =v_2^2=u_0^2-2g{\displaystyle \frac{u_0^2{\sin }^2\alpha }{2g}}=u_0\cos \alpha }$
So, from conservation of momentum
tanθ=mu0cosαmu0cosα=1${\displaystyle \tan \theta ={\displaystyle \frac{m{u}_0\cos \alpha }{m{u}_0\cos \alpha }}=1}$
$$
⇒θ =π/4$\Rightarrow \theta =\pi /4$

Example

Collision of masses in two dimensions

Example: Two small identical spheres each of mass m$m$ are projected slant-wise from the points A$A$ and B$B$ on the ground with equal velocitiesu$u$ and making same angles θ$\theta$ and θ$\theta$ as shown in the figure. They collide with each other at the highest point C$C$ of the common path. If the collision is completely inelastic, how much time after the collision the particles come back to the ground?

Solution:At the highest point, the velocity of the masses will only be horizontal and will be equal to ucosθ$u\cos \theta$.
After the masses stick together, from the formula given in the concept
v=mucosθ−mucosθm+m=0$v={\displaystyle \frac{mu\cos \theta -mu\cos \theta }{m+m}}=0$
So the body will be in free fall starting from rest.
For the vertical motion
12gt2=Hmax=u2sin2θ2g${\displaystyle \frac{1}{2}}g{t}^2=H_{max}={\displaystyle \frac{u^2{\sin }^2\theta }{2g}}$
Solving this, we get t=usinθg$t={\displaystyle \frac{u\sin \theta }{g}}$

Example

Problems on collision combined with splitting of a mass

Example:  A particle is projected with a velocity of 20ms−1$20m{s}^{-1}$ at an angle
600${60}^0$  with the horizontal. At the maximum height it splits into two parts of equal masses. If one part just drops down, what is the velocity of the other part?
Solution:
Velocity at max height =20cos60$=20cos60$
=10$=10$ m/s.
i.e.,
using momentum conservation
10×2m=m×v$10\times 2m=m\times v$
⇒v=20$\Rightarrow v=20$ m/s.

Example

Application of multiple elastic collisions in two-dimensions

Example:
A ball is projected horizontally from top of a 80$80$ m deep well with velocity 10$10$ m/s. Then particle will fall on the bottom at a distance of (all the collisions with the wall are elastic)Solution:
Total time taken by the ball to reach at bottom =2Hg−−−−√=2×8010−−−−−−√=4 sec$=\sqrt{{\displaystyle \frac{2H}{g}}}=\sqrt{{\displaystyle \frac{2\times 80}{10}}}=4sec$
Let time taken in one collision is tcollision$t_{collision}$. Then tcollision×10=7$t_{collision}\times 10=7$ since horizontal velocity is 10 m/s
⇒tcollision=0.7$\Rightarrow t_{collision}=0.7$ sec.
No. of collisions =40.7=557$={\displaystyle \frac{4}{0.7}}=5{\displaystyle \frac{5}{7}}$ (5th collisions from wall B)
Horizontal distance travelled in between 2$2$ successive collisions =7m$=7m$
∴$\therefore$Horizontal distance travelled in 5/7$5/7$ part of collisions =$=$ The 1st collision will be with wall B, 2nd cillision with wall A, 3rd collision with wall B, 4th collision with wall A and  5th collision will be with wall B.  After 5th collision, the ball   will cover a distance of 57×7=5m${\displaystyle \frac{5}{7}}\times 7=5m$ from wall B i.e it will have collision at a distance of 2 m$2m$ from wall A.

Example

Solving Problem involving completely inelastic collision

Example: Three objects A, B and C are kept in a straight line on a frictionless horizontal surface. These have masses m,2m$m,2m$ and m$m$, respectively.The object A moves towards B with a speed 9 m/s and makes an elastic collision with it. Thereafter, B makes completely inelastic collision with C. All motions occur on the same straight line. Find the final speed (in m/s) of the object C.

Solution:After 1st collision
mvA=mv′A+2mv′B$\mathrm{m}{\mathrm{v}}_{\mathrm{A}}=\mathrm{m}{\mathrm{v}}_{\mathrm{A}}^{\prime }+2\mathrm{m}{{\mathrm{v}}^{\prime }}_{\mathrm{B}}$
−1=v′B−v′A0−vA⇒v′B=6m/s$-1={\displaystyle {\displaystyle \frac{{{\mathrm{v}}^{\prime }}_{\mathrm{B}}-{{\mathrm{v}}^{\prime }}_{\mathrm{A}}}{0-{\mathrm{v}}_{\mathrm{A}}}}\Rightarrow {{\mathrm{v}}^{\prime }}_{\mathrm{B}}=6\mathrm{m}/\mathrm{s}}$
After the 2nd collision 
2mv′B=(2m+m)VC⇒vC=23v′B⇒vC=4m/s$2\mathrm{m}{\mathrm{v}}_{\mathrm{B}}^{\prime }=(2\mathrm{m}+\mathrm{m}{)}_{{\mathrm{V}}_{\mathrm{C}}}\Rightarrow {\displaystyle {\mathrm{v}}_{\mathrm{C}}={\displaystyle \frac{2}{3}}{\mathrm{v}}_{\mathrm{B}}^{\prime }\Rightarrow {\mathrm{v}}_{\mathrm{C}}=4\mathrm{m}/\mathrm{s}}$

Example

Problem on perfectly inelastic collision in one dimension

Example: A particle of mass m moving in the direction x$x$ with speed 2v$2v$ is hit by another particle of mass 2m moving in the y$y$ direction with speed v$v$. If the collision is perfectly inelastic, what is the percentage loss in the energy during the collision?
Solution:
as the collision is inelastic the masses will move together 
Assuming the speed of block A and B becomes v1i^+v2j^$v_1\hat{i}+v_2\hat{j}$
Writing momentum equation in x direction
m(2v)+2m(0)=3mv1⇒v1=2v3$m(2v)+2m(0)=3m{v}_1\Rightarrow v_1={\displaystyle \frac{2v}{3}}$
Writing momentum equation in Y direction
m(0)+2m(v)=3mv2⇒v2=2v3$m(0)+2m(v)=3m{v}_2\Rightarrow v_2={\displaystyle \frac{2v}{3}}$
The velocity of both blocks will be 2v3i^+2v3j^${\displaystyle \frac{2v}{3}}\hat{i}+{\displaystyle \frac{2v}{3}}\hat{j}$
The loss in kinetic energy 
12m(2v)2+122m(v)2−123m(2√2v3)2${\displaystyle \frac{1}{2}}m(2v{)}^2+{\displaystyle \frac{1}{2}}2m(v{)}^2-{\displaystyle \frac{1}{2}}3m({\displaystyle \frac{\sqrt{2}2v}{3}}{)}^2$
3mv2−43mv2=53mv2$3m{v}^2-{\displaystyle \frac{4}{3}}m{v}^2={\displaystyle \frac{5}{3}}m{v}^2$
percentage loss53mv23mv2×100=56${\displaystyle \frac{{\displaystyle \frac{5}{3}}m{v}^2}{3m{v}^2}}\times 100=56$ %

Example

Problems on Change in Kinetic Energy

Example: A body starts from rest and is acted on by a constant force. What is the ratio of kinetic energy gained by it in the first five seconds to that gained in the next five second?

Solution: 1) constant   force   →$\to$   constant   acceleration =a$=a$
⇒V5−v0t=a=V5−05=a⇒V5=5a$\Rightarrow {\displaystyle \frac{V_5-v_0}{t}}=a={\displaystyle \frac{V_5-0}{5}}=a\Rightarrow V_5=5a$
⇒K.E5=12mV25=252ma2$\Rightarrow K.E_5={\displaystyle \frac{1}{2}}m{V}_5^2={\displaystyle \frac{25}{2}}m{a}^2$
 2) v10−v0t=a⇒v10=10a⇒K.E10=1002ma2${\displaystyle \frac{v_{10}-v_0}{t}}=a\Rightarrow v_{10}=10a\Rightarrow K.E_{10}={\displaystyle \frac{100}{2}}m{a}^2$
So  K.E  gained  between  5s  and  10s  =K.E10−K.E5$=K.E_{10}-K.E_5$
=752ma2$={\displaystyle \frac{75}{2}}m{a}^2$
K.E  gained  in  5s=252ma2$={\displaystyle \frac{25}{2}}m{a}^2$
⇒Ratio=2575=1:3

BookMarks

Page 1  Page 2  Page 3  Page 4  Page 5  Page 6  Page 7  Page 8  Page 9