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A 3.1g Ball makes a perfect inelastic collision with a second ball which is at rest. If the combined mass moves at a velocity of a quarter the original velocity Of the 3.1g ball, What is the mass of the second ball in grams?
Truck moves with a velocity of 43 m\s and collides with the stationary car with exactly 1 / 5 of its mass. If the two vehicles are locked together, calculate the magnitude of the velocity of their combined masses immediately after the collision?
A 6.8 kg Cannonball is flying at 15.8 m/s [E] when it explodes into two fragments. One 1.4kg fragments (A) Goes off at 8.1 m/s [510 S of E]. What will be the magnitude of the velocity of the second fragment (B) immediately after the explosion? Assume that no mass was lost during the explosion.
To solve these problems, we can use the principle of conservation of momentum. The principle states that the total momentum before a collision is equal to the total momentum after the collision.
1. In the first problem, we have a 3.1g ball colliding with a second ball that is at rest. We are given that the combined mass (which includes the second ball) moves at a velocity that is a quarter of the original velocity of the 3.1g ball.
Let's denote the mass of the second ball as m2. We can set up the conservation of momentum equation as follows:
(m1 * v1) + (m2 * 0) = (m1 + m2) * (v1/4)
Here, m1 is the mass of the 3.1g ball, v1 is its original velocity, and 0 represents the initial velocity of the second ball since it is at rest. We can simplify the equation:
m1 * v1 = (m1 + m2) * (v1/4)
Now, substitute the given values into the equation. Remember to convert the mass of the 3.1g ball to kg:
(0.0031 kg) * (v1) = (0.0031 kg + m2) * (v1/4)
Next, we can isolate m2:
0.0031 kg * v1 = (0.0031 kg + m2) * (v1/4)
0.0031 kg * 4 = 0.0031 kg + m2
0.0124 kg - 0.0031 kg = m2
m2 = 0.0093 kg (or 9.3 g)
Therefore, the mass of the second ball is 9.3 grams.
2. In the second problem, we have a truck moving with a velocity of 43 m/s colliding with a stationary car that has 1/5 of the truck's mass. After the collision, the two vehicles are locked together.
We can again use the conservation of momentum principle to solve this problem. Let's denote the mass of the truck as m1 and the mass of the car as m2. The total momentum before the collision is:
(m1 * v1) + (m2 * 0) = (m1 + m2) * vf
Here, v1 is the initial velocity of the truck, 0 represents the initial velocity of the car since it is stationary, and vf is the final velocity of their combined masses.
We are given that the car has 1/5 of the truck's mass, so we can write:
m2 = (1/5) * m1
Substituting this into the conservation of momentum equation:
(m1 * 43) + (m2 * 0) = (m1 + m2) * vf
(43 * m1) = (m1 + (1/5) * m1) * vf
(43 * m1) = (6/5) * m1 * vf
Now we can simplify and solve for vf:
(43 * 5) = 6 * vf
215 = 6 * vf
vf = 35.833 m/s
Therefore, the magnitude of the velocity of their combined masses immediately after the collision is 35.833 m/s.
3. In the third problem, we have a cannonball flying at 15.8 m/s [E] that explodes into two fragments. One fragment (Fragment A) with a mass of 1.4 kg goes off at 8.1 m/s [510° S of E]. We need to find the magnitude of the velocity of the second fragment (Fragment B) immediately after the explosion, assuming no mass was lost during the explosion.
Again, we can apply the conservation of momentum principle. Denoting the velocity of Fragment B as vb, we can set up the equation:
(ma * va) + (mb * 0) = (ma * 0) + (mb * vb)
Here, ma is the mass of Fragment A, va is its velocity, mb is the mass of Fragment B, and vb is its velocity.
Substituting the given values:
(1.4 kg * 8.1 m/s) + (mb * 0) = (1.4 kg * 0) + (mb * vb)
11.34 kg m/s = 0 + (mb * vb)
11.34 kg m/s = mb * vb
Since we are looking for the magnitude of the velocity, we can ignore the negative sign in front of vb. Therefore:
11.34 kg m/s = mb * |vb|
Now, we need to find the magnitude of vb. To do this, we can square both sides of the equation:
(11.34 kg m/s)^2 = (mb * |vb|)^2
128.4756 kg^2 m^2/s^2 = (mb^2) * (vb^2)
Next, divide both sides by (mb^2):
128.4756 kg^2 m^2/s^2 / mb^2 = vb^2
128.4756 kg m2/s^2 = vb^2
Finally, take the square root of both sides to find the magnitude of vb:
√(128.4756 kg m2/s^2) = vb
11.34 m/s = vb
Therefore, the magnitude of the velocity of the second fragment immediately after the explosion is 11.34 m/s.