thanks Two blocks can collide in a one-dimensional collision. The block on the l
ID: 1414135 • Letter: T
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Two blocks can collide in a one-dimensional collision. The block on the left has a mass of 0.50 kg and is initially moving to the right at 2.4 m/s toward a second block of mass 0.50 kg that is initially at rest. When the blocks collide, a cocked spring releases 1.2 J of energy into the system. (For velocities, use + to mean to the right, - to mean to the left.) What is the velocity of the first block after the collision? What is the velocity of the second block after the collision? Remember that the blocks cannot pass through each other!Explanation / Answer
Let's use the following:
u1 ... speed of block 1 before collision
v1 ... speed of block 1 after collision
v2 ... speed of block 2 after collision
From the conservation of momentums, we get
m1u1 = m1v1 + m2v2
From the conservation of energy + the added 1.2J (which we use only for moving things around - not for heating or scratching the blocks), we get
m1u1²/2 + 1.2 = m1v1²/2 + m2v2²/2
So we have two equations for the unknowns v1 and v2, which we can (try to) solve: From the first equation, we get
v2 = m1/m2*(u1-v1)
Using this in the second equation, we get
m1u1² + 2.4 = m1v1² + (m1*(u1-v1))²/m2
which is a quadratic equation for v1. Its solution gives the speed of v1. Putting this into the previous equation, we get v2.
Using the numbers, the equation to be solved is
0.5*2.4² + 2.4 = 0.5*v1² + 0.25*(2.4-v1)²/0.5
5.28 = 0.5*v1² + 0.5*(2.4-v1)²
10.56 = v1² + 2.4²-4.8*v1+v1²
=> 2 v1² - 4.8v1 - 4.8 = 0
=> v1² - 2.4v1 - 2.4 = 0
=> (v1 - 1.2)^2 = 3.84
=> v1 - 1.2 = +/-1.96
=> v1 = 3.16 m/s
Taking -ve value -
=> v1 = -1.96 + 1.2 = - 0.76 m/s
So,
v2 = m1/m2*(u1-v1) = 0.5/0.5*(2.4 + 0.76) = 3.16 m/s
Only the later solution make sense.
So, v1 = -0.76 m/s
and v2 = 3.16 m/s
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