Answered

Consider two masses m1 and m2 at the top of two frictionless inclined planes. Both masses start from rest at the same height. However, the plane on which m1 sits is at an angle of 30° with the horizontal, while the plane on which m2 sits is at 60°. If the masses are released, which is going faster at the bottom of its plane?

Answer :

Answer:Reach bottom with same velocity

Explanation:

Given

Two masses with [tex]m_1[/tex] and [tex]m_2[/tex] mass respectively

it is given that surface is friction less therefore slipping occurs

Let h be the height from which both masses is released

thus length travel by both the masses is [tex]\frac{h}{\sin 30}[/tex] and [tex]\frac{h}{\sin 60}[/tex]

and acceleration during motion is [tex]g\sin\theta [/tex]

[tex]v^2-u^2=2 a s[/tex]

where [tex]v=final\ velocity[/tex]

[tex]u=initial\ velocity[/tex]

[tex]a=acceleration[/tex]

[tex]s=distance\ traveled[/tex]

[tex]v_1^2-0=2\times g\sin 30\times \frac{h}{\sin 30}[/tex]

[tex]v_1=\sqrt{2gh}[/tex]

For [tex]m_2[/tex]

[tex]v_2^2-u^2=2 a s [/tex]

[tex]v_2^2=2\times g\sin 60 \times \frac{h}{\sin 60}[/tex]

[tex]v_2=\sqrt{2gh}[/tex]

we can see [tex]v_1=v_2[/tex] therefore both masses reach the bottom with common velocity

Other Questions