A particle is projected at angle 60º with speed , from the point ' A ' as shown in the fig. At the same time the wedge is made to move with speed  towards right as shown in the figure. Then the time after which particle will strike with wedge is (g = 10 m/sec2) :

To determine the time after which the particle will strike the wedge, we need to analyze the motion of both the particle and the wedge. 

### Analyzing the Particle's Motion

1. **Initial Conditions:**
   - Initial speed of the particle: \( u \)
   - Angle of projection: \( 60^\circ \)
   - Horizontal component of the initial velocity: \( u_x = u \cos 60^\circ = \frac{u}{2} \)
   - Vertical component of the initial velocity: \( u_y = u \sin 60^\circ = \frac{\sqrt{3}u}{2} \)

2. **Equations of Motion:**
   - Horizontal motion:
     \[
     x_p(t) = \left( \frac{u}{2} \right)t
     \]
   - Vertical motion:
     \[
     y_p(t) = \left( \frac{\sqrt{3}u}{2} \right)t - \frac{1}{2} g t^2
     \]

### Analyzing the Wedge's Motion

The wedge moves with a constant speed \( v \) towards the right. Therefore, its position as a function of time is given by:
\[
x_w(t) = vt
\]

### Condition for Collision

For the particle to strike the wedge, their horizontal positions must be the same at the same time \( t \):
\[
x_p(t) = x_w(t)
\]
Substituting the expressions for \( x_p(t) \) and \( x_w(t) \):
\[
\left( \frac{u}{2} \right)t = vt
\]

### Solving for Time

Simplifying the equation:
\[
\frac{u}{2} = v
\]
\[
u = 2v
\]

Thus, the time \( t \) is the same for both the horizontal positions. Now, we need to determine the time \( t \) in terms of the vertical motion. The particle must be at the same vertical height as the wedge when it strikes.

### Vertical Motion Analysis

The vertical distance \( y_p(t) \) covered by the particle is given by:
\[
y_p(t) = \left( \frac{\sqrt{3}u}{2} \right)t - \frac{1}{2} g t^2
\]

For the particle to strike the wedge, the height must be zero (assuming the wedge is at ground level when struck):
\[
0 = \left( \frac{\sqrt{3}u}{2} \right)t - \frac{1}{2} g t^2
\]

### Simplifying the Equation

\[
\left( \frac{\sqrt{3}u}{2} \right)t = \frac{1}{2} g t^2
\]
\[
\sqrt{3}u t = g t^2
\]
\[
t (\sqrt{3}u - g t) = 0
\]

The non-trivial solution is:
\[
\sqrt{3}u = g t
\]
\[
t = \frac{\sqrt{3}u}{g}
\]

### Final Expression for Time

Using \( u = 2v \):
\[
t = \frac{\sqrt{3} (2v)}{g}
\]
\[
t = \frac{2 \sqrt{3} v}{10}
\]
\[
t = \frac{\sqrt{3} v}{5}
\]

Thus, the time after which the particle will strike the wedge is \( t = \frac{\sqrt{3} v}{5} \).

Do you want more details or have any questions?

### Relative Questions
1. How does the angle of projection affect the range of a projectile?
2. What would be the effect if the wedge had an acceleration instead of a constant speed?
3. Can we solve the problem using energy conservation principles?
4. How would air resistance affect the motion of the particle?
5. What are the components of velocity for a projectile at any point in its trajectory?
6. How does the horizontal displacement of the wedge change if its speed is doubled?
7. Can we use vector analysis to solve this problem?
8. What are the kinematic equations for projectile motion?

### Tip
When solving problems involving both horizontal and vertical motion, it's crucial to separate the motions into their respective components and analyze them independently before combining the results.

Learn how to solve a projectile motion problem where a particle is projected at an angle of 60 degrees with a given speed and strikes a moving wedge. Includes step-by-step analysis of horizontal and vertical motions, using equations of motion for projectiles. Suitable for high school students studying kinematics and vectors.

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