One lump of clay traveling at 14.98 m/s overtakes a second lump traveling at 4.98 m/s. After collision they are stuck together. To the nearest tenth of a m/s what is their common velocity?
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4
If the second lump in the above problem was moving to the left before the collision what would be the answer?
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5
A ball strikes a second ball at rest in an elastic collision. If the mass of the first ball is 27 kg and the mass of the second is 1.7 kg, to the nearest tenth of a m/s, if the first ball is initially traveling at 11.7 m/s, what is its velocity after the collision?
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6
In the preceeding problem what is the velocity of the second ball after collision?
To solve these problems, we can use the principles of conservation of momentum and conservation of kinetic energy.
1) For the first problem, we have two lumps of clay with different velocities before the collision. After the collision, they stick together, indicating a completely inelastic collision. In an inelastic collision, the total momentum is conserved.
The momentum (p) of each lump can be calculated using the formula: p = mass * velocity.
Given:
Mass of first lump = mass1 = unknown
Velocity of first lump = velocity1 = 14.98 m/s
Mass of second lump = mass2 = unknown
Velocity of second lump = velocity2 = 4.98 m/s
Let's label the final velocity (common velocity) after the collision as V.
To solve for the common velocity, we can set up the following equation:
(mass1 * velocity1) + (mass2 * velocity2) = (mass1 + mass2) * V
Since we don't know the mass of each lump, we can assign arbitrary variables such as m1 and m2.
Let's solve it step by step:
Step 1: Let mass1 = m1 and mass2 = m2.
Step 2: Rewrite the equation:
(m1 * 14.98) + (m2 * 4.98) = (m1 + m2) * V
Step 3: Rearrange the equation to solve for V:
(m1 * 14.98) + (m2 * 4.98) = (m1 + m2) * V
14.98m1 + 4.98m2 = V(m1 + m2)
14.98m1 + 4.98m2 = Vm1 + Vm2
Step 4: Combine like terms:
10.00m1 = (V - 4.98)m2
10.00m1 / m2 = V - 4.98
Step 5: Solve for V:
V = (10.00m1 / m2) + 4.98
Since we don't have the exact values for the masses, we can't determine the specific common velocity. However, we can use the given values to estimate the answer to the nearest tenth of a m/s.
2) In this problem, we still have an inelastic collision between two lumps of clay. However, the second lump is moving to the left before the collision.
Using the same approach as in problem 1, but considering the opposite direction of the second lump's velocity, we can calculate the common velocity after the collision.
3) For the third problem, we have two balls colliding in an elastic collision. In an elastic collision, both momentum and kinetic energy are conserved.
Given:
Mass of first ball = m1 = 27 kg
Velocity of first ball before the collision = velocity1 = 11.7 m/s
Mass of second ball = m2 = 1.7 kg
Velocity of second ball before the collision = 0 m/s (at rest)
To find the velocity of the first ball after the collision, we can use the equation for conservation of momentum:
(m1 * velocity1) + (m2 * 0) = (m1 * V1) + (m2 * V2)
Since the second ball is at rest (velocity2 = 0), it does not contribute to the momentum after the collision. Therefore, the equation simplifies to:
(m1 * velocity1) = (m1 * V1)
Dividing both sides by m1:
velocity1 = V1 (velocity of the first ball before the collision equals the velocity of the first ball after the collision)
Therefore, the velocity of the first ball after the collision is 11.7 m/s.
4) In the preceding problem, since the second ball is at rest (velocity2 = 0) in an elastic collision, it will gain the entire momentum from the first ball. Therefore, the velocity of the second ball after the collision will be the same as the velocity of the first ball before the collision, which is 11.7 m/s.