A body of mass m moving vertically with speed 3 m/s hits a smooth fixed inclined plane and rebounds with a velociw vf in the horizontal direction. If angle of inclined plane is 30°, the velocity of will be
1/root3 m/s
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√3
To find the velocity of the body after it rebounds off the inclined plane, we can use the principles of conservation of energy and momentum.
First, let's consider the initial state of the body when it hits the inclined plane. The body has a mass m and is moving vertically with a speed of 3 m/s. Therefore, the initial kinetic energy (KEi) of the body can be calculated using the formula:
KEi = (1/2) * m * (3^2)
Next, let's consider the final state of the body after it rebounds off the inclined plane. The body will be moving in the horizontal direction with velocity vf. The final kinetic energy (KEf) of the body can be calculated using the formula:
KEf = (1/2) * m * vf^2
Since the inclined plane is smooth, we can assume there is no loss of mechanical energy during the collision. Therefore, the initial potential energy (PEi) will be equal to the final kinetic energy:
PEi = KEf
The initial potential energy of the body can be calculated using the formula:
PEi = m * g * h
where g is the acceleration due to gravity and h is the height of the inclined plane. In this case, h can be calculated using the formula:
h = (3^2) / (2 * g)
Now, we can substitute the values into the equation and solve for vf:
m * g * h = (1/2) * m * vf^2
Simplifying this equation, we get:
vf^2 = 2 * g * h
vf = sqrt(2 * g * h)
Finally, substitute the given values, where angle of the inclined plane is 30°, into the equation and solve for vf:
vf = sqrt(2 * 9.8 * (3^2) / (2 * 9.8))
vf = sqrt(2 * 9.8 * 9 / 2)
vf = sqrt(2 * 9.8 * 4.5)
vf = sqrt(88.2)
vf ≈ 9.4 m/s
Therefore, the velocity of the body after it rebounds off the inclined plane will be approximately 9.4 m/s in the horizontal direction.