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.
v1=
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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
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
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..
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!
can you elaborate..
Use conservation of momenta
you should get
0 = -m2v2 + m1v1 assuming that the block 2 will move to the left
now that gets you to a great relationship between v1 and v2, i suggest get v2 in terms of v1
so v2 = m1v1/m2
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
To find the speed of the cube (v1) as it leaves the block, we can apply the principle of conservation of mechanical energy.
At the top of the circular track, the cube only possesses gravitational potential energy (m1 * g * h) since it is at rest. Here, h is the height at which the cube is placed.
As the cube slides down the track and reaches the bottom, it loses this gravitational potential energy and gains an equal amount of kinetic energy. At the bottom of the track, the kinetic energy of the cube can be represented as (1/2 * m1 * v1^2), where v1 is the speed of the cube as it leaves the block.
Since the track is frictionless, the total mechanical energy of the system (cube + block) is conserved throughout the motion. Therefore, the gravitational potential energy lost by the cube is gained as kinetic energy by the cube-block system.
The total mechanical energy of the system is the sum of the gravitational potential energy and the kinetic energy of the block:
m1 * g * h = (1/2 * (m1 + m2) * v2^2) + m2 * g * h
Here, v2 is the speed of the block as the cube leaves it at the bottom.
Since the block is at rest initially and remains at rest after the cube leaves it, v2 is zero. Therefore, the equation simplifies to:
m1 * g * h = (1/2 * (m1 + m2) * v1^2)
Now, we can solve for v1. Rearranging the equation:
v1^2 = (2 * m1 * g * h) / (m1 + m2)
Substituting the given values:
v1^2 = (2 * 1.0 kg * 10.0 m/s^2 * h) / (1.0 kg + 4.0 kg)
v1^2 = (20.0 * h) / 5.0 (m^2/s^2 conveniently cancels out)
v1^2 = 4.0 * h (m^2/s^2 conveniently cancels out)
To find v1, we need to know the value of h. Once we have h, we can substitute it into the equation to find v1.
Note: The value of h is not provided in the question or figure. Hence, without the value of h, we cannot determine the exact value of v1.