A particle travels horizontally between two parallel walls separated by 18.4m. It moves towards the opposing wall at a constant rate of 7.2m/s. Also, it has an acceleration in the direction parallel to the walls of 2 m/s^2.
a) What will its speed be when it hits the wall?
b) At what angle with the wall will the particle strike?
To find the answers to these questions, we need to use the equations of motion.
a) To find the speed of the particle when it hits the wall, we can use the equation:
\[v_f = v_i + at\]
where:
- \(v_f\) is the final velocity (speed).
- \(v_i\) is the initial velocity.
- \(a\) is the acceleration.
- \(t\) is the time.
In this case, the initial velocity is 7.2 m/s, the acceleration is 2 m/s\(^2\), and we want to find the final velocity. Since the particle is moving towards the opposing wall, the final velocity will be negative.
Let's first find the time it takes for the particle to hit the wall. We can use the equation:
\[d = v_i t + \frac{1}{2} a t^2\]
where:
- \(d\) is the distance traveled.
- \(v_i\) is the initial velocity.
- \(a\) is the acceleration.
- \(t\) is the time.
In this case, the distance traveled is the distance between the walls, 18.4 m.
Plugging in the values, we get:
\[18.4 = 7.2 t + \frac{1}{2} \cdot 2 \cdot t^2\]
This is a quadratic equation. Solving it will give us the time it takes for the particle to hit the wall.
b) To find the angle at which the particle strikes the wall, we need to calculate the angle of velocity vector v with respect to the wall.
Let θ be the angle between the velocity vector and the wall. The tangent of θ is given by:
\[tan(\theta) = \frac{v_y}{v_x}\]
where:
- \(v_y\) is the vertical component of velocity.
- \(v_x\) is the horizontal component of velocity.
To find the vertical and horizontal components of velocity, we can use the equations:
\[v_x = v \cdot cos(\theta)\]
\[v_y = v \cdot sin(\theta)\]
where:
- \(v\) is the magnitude of the velocity vector.
To calculate the magnitude of the velocity vector, we can use the final velocity obtained from part (a).
Now, we can substitute the values of \(v_x\), \(v_y\), and \(v\) into the equation for the tangent of θ to find the angle θ.
Let's apply these equations to find the answers to the questions.
The constant 7.2 m/s velocity is the component perpendicular to the walls. Call it Vx. The particle only accelerates in the direction parallel to the walls.
The time it takes to cross from one wall t the other is
D/Vx = (18.4 m)/(7.2 m/s) = 2.56 s
In that time, it will have acquired a vertical velocity component of
Vy = (1/2)*2*(2.56)^2 = 6.53 m/s
The speed when it hits the wall is
V = sqrt[(6.53)^2 + (7.2)^2] = 9.72 m/s
The velocity has an an angle at impact, with respect to the horizontal, of
arctan (Vy/Vx) = arctan(.907) = 42.2 degrees