A 4 kg mass and a 9 kg mass are being held at rest against a compressed spring on a frictionless surface. When the masses are released, the 4 kg mass moves to the east with a speed of 2m/s. What is the velocity of the 9 kg after the are released?
This is a inelastic equation, so we use (m1+m2)vf = m1v1+m2v2, where m1 is 4 kg, m2= 9kg. The start with no velocity so Vf=0, meaning the left side is 0.
m1v1=4 times 2 =8, We subtract to the left side and divide by m2, 9kg. v2 = 8/9 heading west.
Well, I hate to be the bearer of bad news, but if you've got a 4 kg mass moving east at 2 m/s, and a 9 kg mass, I'm sorry to say that the 9 kg mass will not be moving. In fact, it will be sitting there, minding its own business, wondering why the 4 kg mass is hogging all the limelight. Poor 9 kg mass, always left in the dust. So to answer your question, the velocity of the 9 kg mass is a big fat zero.
To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum before an event is equal to the total momentum after the event, assuming there are no external forces acting on the system.
1. Calculate the initial momentum of the system.
The initial momentum is given by the sum of the individual momenta of the 4 kg and 9 kg masses.
Momentum of the 4 kg mass (p1) = mass (m1) × velocity (v1) = 4 kg × 2 m/s = 8 kg·m/s
Momentum of the 9 kg mass (p2) = mass (m2) × velocity (v2)
2. Since the masses are at rest initially and released from a compressed spring (which implies the conservation of mechanical energy), the momentum of the system is conserved.
Therefore, the total momentum before the release is equal to the total momentum after the release:
Total initial momentum = Total final momentum
p1 + p2 = p1' + p2'
Note: The primes (') indicate the final velocities.
3. The momentum of the 4 kg mass after the release is given: p1' = m1 × v1' where v1' is the final velocity of the 4 kg mass, which is 2 m/s to the east.
p1' = 4 kg × 2 m/s = 8 kg·m/s
Substitute p1' = 8 kg·m/s in the equation p1 + p2 = p1' + p2':
8 kg·m/s + p2 = 8 kg·m/s + p2'
4. Since the total momentum before the release is equal to the total momentum after the release, p2 = p2'.
Substitute p2 = p2' in the equation 8 kg·m/s + p2 = 8 kg·m/s + p2':
8 kg·m/s + p2 = 8 kg·m/s + p2
The equation is satisfied by any value of p2. Hence, the velocity of the 9 kg mass after release could be any value.
In conclusion, the velocity of the 9 kg mass after the release can be any value.
To find the velocity of the 9 kg mass after release, we can apply the principle of conservation of momentum. According to this principle, the total momentum before the release of the masses should be equal to the total momentum after the release.
The momentum of an object is defined as the product of its mass and velocity. Let's represent the velocity of the 4 kg mass as v1 and the velocity of the 9 kg mass as v2.
The total initial momentum before release is 0 since both masses are at rest. After release, the 4 kg mass moves with a velocity of 2 m/s to the east. Therefore, the momentum of the 4 kg mass is given by:
momentum of 4 kg mass = mass of 4 kg * velocity v1
momentum of 4 kg mass = 4 kg * 2 m/s = 8 kg·m/s
Since the total initial momentum is 0, this means the momentum of the 9 kg mass after release should be equal in magnitude but in the opposite direction. Therefore:
momentum of 9 kg mass = -8 kg·m/s
Now, we can find the velocity of the 9 kg mass. Using the definition of momentum:
momentum of 9 kg mass = mass of 9 kg * velocity v2
-8 kg·m/s = 9 kg * v2
Solving for v2:
v2 = (-8 kg·m/s) / 9 kg
v2 ≈ -0.89 m/s
Therefore, the velocity of the 9 kg mass after release is approximately -0.89 m/s, indicating it moves to the west.