Question

A particleA is dropped from a point P at t=0. At the same time another particleB is thrown from a point O which is 2m to the right and 10m below P , with velocity v making an angle @ with horizontal.
B collides with the particle A . If the two particles collide 2 sec after they start,find v.
(Acceleration due to gravity is 10m/s in negative y direction. Point O in not necessarily on ground.)
Ans- (root)26

Please help me solve this using x and y components of v. And also where would they collide?

Answer 1

Answer on Question #74040 Physics / Classical Mechanics

A particle $A$ is dropped from a point $P$ at $t = 0$. At the same time another particle $B$ is thrown from a point $O$ which is $2\mathrm{m}$ to the right and $10\mathrm{m}$ below $P$, with velocity $v$ making an angle $\theta$ with horizontal. $B$ collides with the particle $A$. If the two particles collide 2 sec after they start, find $v$. (Acceleration due to gravity is $10\mathrm{m/s}$ in negative y direction. Point $O$ in not necessarily on ground.)

Solution:

Let the origin is at point $O$. The $y$ axis is directed upward. The $x$ axis is directed to the left. So, the equations of motion of particles $A$ and $B$

$$y_A(t) = 10 - \frac{g t^2}{2}$$ $$x_A(t) = 2$$ $$y_B(t) = v \sin \theta \quad t - \frac{g t^2}{2}$$ $$x_B(t) = v \cos \theta \quad t$$

When particles collide

$$x_A(t) = x_B(t) \quad \text{and} \quad y_A(t) = y_B(t), \quad t = 2\mathrm{s}$$

Thus

$$2 = 2 v \cos \theta \Rightarrow v \cos \theta = 1$$ $$10 - \frac{10 \times 2^2}{2} = 2 v \sin \theta - \frac{10 \times 2^2}{2} \Rightarrow v \sin \theta = 5$$

Finally

$$v^2 \cos^2 \theta + v^2 \sin^2 \theta = 1^2 + 5^2$$ $$v^2 = 26$$ $$v = \sqrt{26} \mathrm{m/s}$$

Answer: $v = \sqrt{26} \, \mathrm{m/s}$

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