A small block of mass 55 g is released from rest at the top of a curved friction
ID: 1284312 • Letter: A
Question
A small block of mass 55 g is released from rest at the top of a curved frictionless wedge of mass 314 g which sits on a frictionless horizontal surface as shown. The acceleration due to gravity is 9.81 m/s2. Right is the positive direction.
When the block leaves the wedge, its velocity is 2.52 m/s to the right.
What is the velocity vb of the wedge after the block reaches the horizontal surface?
Answer in units of m/s
What is the height h of the wedge? Answer in units of m
I found the velocity -0.441401 but I can't find the height of the wedge.
A small block of mass 55 g is released from rest at the top of a curved frictionless wedge of mass 314 g which sits on a frictionless horizontal surface as shown. The acceleration due to gravity is 9.81 m/s^2. Right is the positive direction. What is the velocity vb of the wedge after the block reaches the horizontal surface? Answer in units of m/s What is the height h of the wedge? Answer in units of m I found the velocity -0.441401 but I can't find the height of the wedge. When the block leaves the wedge, its velocity is 2.52 m/s to the right.Explanation / Answer
The initial energy of the system must equal the final energy of the system. Initially, the wedge is not moving and is resting on the ground. Therefore, it has no energy. The block is not moving but is at the top of the wedge. Therefore, it has potential energy. The potential energy of the block must equal the total kinetic energy of the wedge and block once the block reaches the ground.
mbgh=12mbv2b+12mwv2w
.423(9.81)h= .5(.053)(3.64^2)+.5(.370)(-(.521^2)
.772m