Question

A ball of mass m is released from the top of an inclined plane of angle ø. Its strikes a rigid wall at a distance 3l/4 from top elastically. The impulse imparted to ball by rigid wall is
(1) m√3/2gh
(2)m√3gh
(3)2m√3gh
(4)m√6gh
(l= length of hypotenuse of plank)
(h = height of inclined plane)

Answer 1

Answer on Question #50068, Physics, Mechanics | Kinematics | Dynamics

A ball of mass $m$ is released from the top of an inclined plane of angle $\phi$. Its strikes a rigid wall at a distance 3/4 from top elastically. The impulse imparted to ball by rigid wall is

(1) mV3/2gh

(2) mV3gh

(3) 2mV3gh

(4) mV6gh

(l= length of hypotenuse of plank)

(h = height of inclined plane)

Solution:

The impulse of force can be extracted and found to be equal to the change in momentum of an object provided the mass is constant:

$$Impulse = m \Delta v$$

The change in momentum is

$$\Delta v = v_2 - v_1$$

When the strike is elastically

$$v_1 = -v_2 = -v$$

The total mechanical energy in any isolated system of objects remains constant if the objects interact only through conservative forces:

$$\frac{1}{2} m v^2 = m g h_1$$

Thus,

$$v = \sqrt{2 g h_1}$$

Where

$$\frac{h_1}{h} = \frac{\frac{3}{4} l}{l}$$

Hence

$$h_1 = \frac{3}{4} h$$

Impulse is

$$= 2 m v = 2 m \sqrt{\frac{2 g \cdot 3}{4} h} = m \sqrt{\frac{4 \cdot 3 \cdot g h}{2}} = m \sqrt{6 g h}$$

Answer: (4) mV6gh

http://www.AssignmentExpert.com/