Two blocks of masses M and 3M are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them.
A cord initially holding the blocks together is burned; after this the block of mass 3M moves to the right with a speed of v1=2.48 m/s. What is the velocity of the block of mass M?
The force on each end of the spring is the same, the impulse is the same, the momentums are the same.
To solve this question, we can use the principle of conservation of momentum. The initial momentum before the cord is burned should be equal to the final momentum after the cord is burned.
Let's denote the initial velocities of the blocks as v_M and v_3M for the block of mass M and 3M, respectively.
Since the blocks are initially at rest, the total initial momentum is zero:
initial momentum = m_M * v_M + m_3M * v_3M = 0
After the cord is burned, the two blocks move separately, and we know the velocity of the block with mass 3M is v_3M = 2.48 m/s.
The final momentum is given by:
final momentum = m_M * v_M' + m_3M * v_3M
To find v_M', the velocity of the block with mass M after the cord is burned, we need to rearrange the equation and solve for v_M':
m_M * v_M' = -m_3M * v_3M
v_M' = (-m_3M / m_M) * v_3M
Let's substitute the given values:
v_M' = (-3M / M) * 2.48 m/s
v_M' = -7.44 m/s
Therefore, the velocity of the block with mass M after the cord is burned is -7.44 m/s (moving to the left).
To find the velocity of the block of mass M, we can use the principle of conservation of momentum. According to this principle, the total momentum before an event is equal to the total momentum after the event, assuming no external forces are acting on the system.
In this scenario, the system consists of the two blocks attached by the spring. Initially, they are at rest, so the total momentum before the cord is burned is zero.
After the cord is burned, the block of mass 3M moves to the right with a speed of v1 = 2.48 m/s. We need to find the velocity of the block of mass M.
Let's assume the velocity of the block of mass M is v2.
Since both blocks are initially at rest, the initial momentum of the system is zero: 0 = (0 kg) * v_initial
The final momentum of the system is given by the sum of the individual momenta of both blocks:
final momentum = (mass of M) * v2 + (mass of 3M) * v1
According to the principle of conservation of momentum, the initial and final momenta are equal:
0 = (mass of M) * v2 + (mass of 3M) * v1
Now we can substitute the given values into the equation and solve for v2:
0 = M * v2 + 3M * v1
0 = M * v2 + 3M * 2.48
0 = M * v2 + 7.44M
Now we can factor out M:
0 = M * (v2 + 7.44)
Dividing both sides of the equation by M:
0 = v2 + 7.44
Rearranging the equation:
v2 = -7.44 m/s
Therefore, the velocity of the block of mass M is -7.44 m/s. Note that the negative sign indicates that the block is moving to the left.