A small cube of mass m1= 1.0 kg slides down a circular and frictionless track of radius R= 0.6 m cut into a large block of mass m2= 4.0 kg as shown in the figure below. The large block rests on a horizontal and frictionless table. The cube and the block are initially at rest, and the cube m1 starts from the top of the path. Find the speed of the cube v1 as it leaves the block. Take g= 10.0 m/s2. Enter your answer in m/s.

Question

v1=

Answer 1

You must be taking a test.

If you really want help, you need to indicate what you do or don't understand about each of these.

Answer 2

The center of mass of the system will not move in the horizontal plane.

The potential energy lost by the little mass is equal to the total kinetic energy gained by the system

Answer 3

if i make a simple conversion from PE to KE an then velocity it gives me a red cross :( but i don see ho to fit the bigger block into thepicture

Answer 4

I assumed that Normal force exerted by the sliding block on track-block will give that track-block some velocity over time whcih it takes for the sliding block to reach the bottom of the track. So I found(but not sure about it) that Normal Force = sin (Theta) * m*g - it is the function of Theta. I found the impulse which receive track-block as integral from 0 to 90 degrees from Normal force. And from that value I found velocity of track-block m_2*v_2 = Integral of Normal Force. And assuming this speed as a loss in energy in sliding block was trying to find velocity of sliding vlock from conservation of energy... But failed, because obtain an number which is less than zero..

Answer 5

1) concervation of momentum
(p=mv) check what concervation means again,
then you will write a very simple equality
2)m1gh = 0.5m1v1²+0.5m2v2²

thanks to 1) substitute in 2) and you have your formula with only v1

Good luck!

Answer 6

can you elaborate..

Answer 7

Use conservation of momenta

you should get

0 = -m2v2 + m1v1 assuming that the block 2 will move to the left

Answer 8

now that gets you to a great relationship between v1 and v2, i suggest get v2 in terms of v1

so v2 = m1v1/m2

Answer 9

now use conservation of energy "assuming no energy is lost due to this process"

hence you should have

Potential of m1 = Kinetic m1 + Kinetic m2

or

to elaborate

m1gR = 1/2(m1)*(v1)^2 + 1/2(m2)*(v2)^2

given all those formula, you should be able to find an expression for v1 :)
Goodluck