Physics > Motion in Two Dimension > 6.0 Relative motion between two projectiles
Motion in Two Dimension
1.0 Introduction
2.0 Projectile motion
3.0 Ground to ground projectile motion
3.1 Maximum height
3.2 Time of flight
3.3 Range
3.4 Trajectory of a projectile
3.5 Summary
3.6 Solved Examples
4.0 Projectile thrown parallel to the horizontal
5.0 Projectile on an inclined plane
6.0 Relative motion between two projectiles
6.1 Solved examples
3.2 Time of flight
3.3 Range
3.4 Trajectory of a projectile
3.5 Summary
3.6 Solved Examples
Question: A particle $A$ is projected with an initial velocity of $60 \ m/s$ at an angle of $30^\circ $ to the horizontal. At the same time a second particle $B$ is projected in opposite direction with an initial speed of $50\ m/s$ from a point at a distance of $100\ m$ from $A$. If the particle collides in air,
(a). Find the angle of projection $\alpha$ of particle $B$.
(b). Find the time after which the collision takes place.
(Take $g = 10m/{s^2}$)
Solution:
Velocities of the particles are,
${\overrightarrow u _A} = 60\left( {\cos 30^\circ \widehat i + \sin 30^\circ \widehat j} \right)$
${\overrightarrow u _B} = 50\left( { - \cos \alpha \widehat i + \sin \alpha \widehat j} \right)$
So, velocity of particle $A$ relative to particle $B$ (observer) is,
$${\overrightarrow u _{AB}} = {\overrightarrow u _A} - {\overrightarrow u _B}$$$${\overrightarrow u _{AB}} = 60\left( {\cos 30^\circ + \sin 30^\circ \hat j} \right) - 50\left( { - \cos \alpha \hat i + \sin \alpha \hat j} \right)$$$${\overrightarrow u _{AB}} = \left( {60\cos 30^\circ + 50\cos \alpha } \right)\widehat i + \left( {60\sin 30^\circ - 50\sin \alpha } \right)\hat j$$
Acceleration of the particles are,
${\overrightarrow a _A} = - g\widehat j$
${\overrightarrow a _B} = - g\widehat j$
So, acceleration of particle $A$ relative to particle $B$ (observer) is,
$${\overrightarrow a _{AB}} = {\overrightarrow a _A} - {\overrightarrow a _B}$$$${\overrightarrow a _{AB}} = - g\widehat j - \left( { - g\widehat j} \right)$$$${\overrightarrow a _{AB}} = 0$$
(a).
When particle $B$ is observer
For the particle to collide, particle $A$ should move along the line joining particle $A$ and $B$ i.e. along $x-$ axis.
Therefore velocity along $y-$ axis will be zero. Mathematically,
$${\overrightarrow u _{AB}}_y = 0$$or $$\left( {60\sin 30^\circ - 50\sin \alpha } \right) = 0$$$$\sin \alpha = \frac{{60\sin 30^\circ }}{{50}}$$$$\sin \alpha = \frac{3}{5}$$$$\alpha = {\sin ^{ - 1}}\left( {\frac{3}{5}} \right)$$
$${\overrightarrow u _{AB}}_y = 0$$or $$\left( {60\sin 30^\circ - 50\sin \alpha } \right) = 0$$$$\sin \alpha = \frac{{60\sin 30^\circ }}{{50}}$$$$\sin \alpha = \frac{3}{5}$$$$\alpha = {\sin ^{ - 1}}\left( {\frac{3}{5}} \right)$$
(b). Time taken by the particles to collide is,
$$t = \frac{d}{{{u_{A{B_x}}}}}$$$$t = \frac{{100}}{{\left( {60\cos 30^\circ + 50\cos \alpha } \right)}}$$$$t = \frac{{100}}{{\left( {60 \times \frac{{\sqrt 3 }}{2} + 50 \times \frac{4}{5}} \right)}}$$$$t = \frac{{10}}{{\left( {3\sqrt 3 + 4} \right)}}$$
$$t = \frac{d}{{{u_{A{B_x}}}}}$$$$t = \frac{{100}}{{\left( {60\cos 30^\circ + 50\cos \alpha } \right)}}$$$$t = \frac{{100}}{{\left( {60 \times \frac{{\sqrt 3 }}{2} + 50 \times \frac{4}{5}} \right)}}$$$$t = \frac{{10}}{{\left( {3\sqrt 3 + 4} \right)}}$$
Question: A particle is thrown horizontally with velocity $10\ m/s$ from an inclined plane which is also moving with an acceleration of $10\ m/s^2$ vertically upward. Find out the time after which it lands on the inclined plane?
Solution:
Kinematic variable of all the particle are written as,
${\overrightarrow u _P} =-u \widehat i= - 10\widehat i$
${\overrightarrow a _P} =-g \widehat j= - 10\widehat j$
${\overrightarrow u _w} = 0$
${\overrightarrow a _w} = + a\widehat j = + 10\widehat j$So, let the observer be on the wedge. Therefore, we will take velocities, acceleration, and displacement relative to the wedge.
$${\overrightarrow u _{Pw}} = {\overrightarrow u _P} - {\overrightarrow u _w}$$$${\overrightarrow u _{Pw}} = - u\widehat j$$and$${\overrightarrow a _{Pw}} = {\overrightarrow a _P} - {\overrightarrow a _w}$$$${\overrightarrow a _{Pw}} = - g\widehat j - \left( a \right)\widehat j$$$${\overrightarrow a _{Pw}} = - \left( {g + a} \right)\widehat j$$
Now it will be like simple projectile on an inclined plane.
$X$ Horizontal motion | $Y$ Vertical motion |
| ${\overrightarrow u _X} = - u \cos \alpha$ | ${\overrightarrow u _Y} = +u \sin \alpha$ |
| ${\overrightarrow a _X} = -(g+a) \sin \alpha$ | ${\overrightarrow a _Y} = - (g+a) \cos \alpha$ |
When the particle returns back to the inclined plance. (i.e displacement becomes zero)
${\overrightarrow s _Y} = 0$
${\overrightarrow u _Y} = + u\sin \alpha = 10\sin 30^\circ $
${{\vec a}_Y} = - (g + a)\cos \alpha = - (10 + 10)\cos 30^\circ $
${\overrightarrow u _Y} = + u\sin \alpha = 10\sin 30^\circ $
${{\vec a}_Y} = - (g + a)\cos \alpha = - (10 + 10)\cos 30^\circ $
$t=?$
We can write the kinematic equation relating displacement and time as,
$${\overrightarrow s _Y} = {\overrightarrow u _Y}t + \frac{1}{2}{\overrightarrow a _Y}{t^2}$$$$0 = u\sin \alpha t + \frac{1}{2}\left[ { - (g + a)} \right]{t^2}$$$$t = \frac{{2u\sin \alpha }}{{(g + a)\cos \alpha }}$$or$$t = \frac{{2 \times 10 \times \sin 30^\circ }}{{(10 + 10)\cos 30^\circ }}$$$$t = \frac{{20 \times \frac{1}{2}}}{{20 \times \frac{{\sqrt 3 }}{2}}}$$$$t = \frac{1}{{\sqrt 3 }}s$$
$${\overrightarrow s _Y} = {\overrightarrow u _Y}t + \frac{1}{2}{\overrightarrow a _Y}{t^2}$$$$0 = u\sin \alpha t + \frac{1}{2}\left[ { - (g + a)} \right]{t^2}$$$$t = \frac{{2u\sin \alpha }}{{(g + a)\cos \alpha }}$$or$$t = \frac{{2 \times 10 \times \sin 30^\circ }}{{(10 + 10)\cos 30^\circ }}$$$$t = \frac{{20 \times \frac{1}{2}}}{{20 \times \frac{{\sqrt 3 }}{2}}}$$$$t = \frac{1}{{\sqrt 3 }}s$$
