A small block of mass `m_(1)` is released from rest at the top of a curve-shaped, frictionless wedge of mass `m_(2)` which sits on a frictionless horizontal surface as shown. When the block leaves the wedge its velocity is measured to be 4.00 m/s to the right as shown in the figure. if the mass of the block is doubled to becomes `2m_(1)`, what can be said about the speed with which it leaves the wedge ?

Options

A. Its speed is less than 4.00 m/s
B. Its speed is equal to 4.00 m/s
C. Its speed is greter than 4.00 m/s
D. No enough information is given.

Answer

A

Explanation

`m_(1)gh=1/2 muV^(2)_(rel) =1/2 (m_(1)m_(2))/(m_(1)+m_(2)) v^(2)_(rel)`
`v_(rel)=sqrt(2gh(1+(m_(1))/(m_(2))))`
`v_(1)=v_(1c)=(mu)/(m_(1)) v_(rel)`
`=(m_(2))/(m_(1)+m_(2))xxsqrt(2gh(m_(2)+m_(1))/(m_(2)))`
=`sqrt(2gh((m_(2))/(m_(1)+m_(2))))`