A 5.00 g object moving to the right at +20.0 cm/s makes an elastic head-on collision with a 10.0 g object that is initially at rest.
(a) Find the velocity of each object after the collision.
5.00 g object cm/s
10.0 g object cm/s
(b) Find the fraction of the initial kinetic energy transferred to the 10.0 g object.
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thankyou it was really helpful
(a) After the elastic head-on collision, the 5.00 g object and the 10.0 g object will exchange their velocities.
So, the 5.00 g object will come to rest and the 10.0 g object will move to the right.
The velocity of the 5.00 g object after the collision will be 0 cm/s.
The velocity of the 10.0 g object after the collision will be +20.0 cm/s.
(b) To find the fraction of the initial kinetic energy transferred to the 10.0 g object, we need to calculate the kinetic energy before and after the collision.
The initial kinetic energy is given by:
KE_initial = (1/2) * m1 * v1^2 + (1/2) * m2 * v2^2
Where m1 and v1 are the mass and velocity of the 5.00 g object, and m2 and v2 are the mass and velocity of the 10.0 g object before the collision.
Since the 5.00 g object is moving to the right at +20.0 cm/s and the 10.0 g object is initially at rest, we have:
m1 = 5.00 g = 0.005 kg
v1 = +20.0 cm/s
m2 = 10.0 g = 0.01 kg
v2 = 0 cm/s
Plugging these values into the equation, we get:
KE_initial = (1/2) * 0.005 kg * (20.0 cm/s)^2 + (1/2) * 0.01 kg * (0 cm/s)^2
Simplifying, we find:
KE_initial = 0.005 kg * 200.0 cm^2/s^2 + 0.01 kg * 0 cm^2/s^2
KE_initial = 2.5 J
The final kinetic energy of the 10.0 g object is given by:
KE_final = (1/2) * m2 * v_final^2
Where v_final is the velocity of the 10.0 g object after the collision.
Plugging in the values, we find:
KE_final = (1/2) * 0.01 kg * (20.0 cm/s)^2
Simplifying, we get:
KE_final = 2 J
To find the fraction of the initial kinetic energy transferred to the 10.0 g object, we divide the final kinetic energy by the initial kinetic energy and multiply by 100%:
% = (KE_final / KE_initial) * 100%
Plugging in the values, we find:
% = (2 J / 2.5 J) * 100%
% = 80%
So, the fraction of the initial kinetic energy transferred to the 10.0 g object is 80%.
Clown Bot says: "What a generous exchange of energy! The 10.0 g object got quite the ego boost from the collision."
To solve this problem, we can use the principles of conservation of linear momentum and kinetic energy.
(a) First, let's calculate the initial momentum of the system before the collision.
The momentum of an object is given by the product of its mass and velocity.
For the 5.00 g object:
Momentum1 = mass1 * velocity1
= (5.00 g) * (+20.0 cm/s) [Since the object is moving to the right, its velocity is positive.]
For the 10.0 g object initially at rest:
Momentum2 = mass2 * velocity2
= (10.0 g) * (0 cm/s) [Since the object is initially at rest, its velocity is zero.]
The total initial momentum of the system is the sum of the individual momenta:
Initial momentum = Momentum1 + Momentum2
(b) According to the principle of conservation of linear momentum, the total momentum of the system remains constant before and after the collision.
After the collision, let's assume that the 5.00 g object moves to the left and the 10.0 g object moves to the right.
Let's label the post-collision velocity of the 5.00 g object as v1 and the velocity of the 10.0 g object as v2.
Applying the principle of conservation of linear momentum, the total momentum after the collision is also the sum of the individual momenta of the objects.
For the 5.00 g object:
Momentum1' = mass1 * velocity1'
= (5.00 g) * (-v1) [Since the object is moving to the left, its velocity is negative.]
For the 10.0 g object:
Momentum2' = mass2 * velocity2'
= (10.0 g) * (+v2) [Since the object is moving to the right, its velocity is positive.]
The total momentum after the collision is:
Final momentum = Momentum1' + Momentum2'
Since the principle of conservation of linear momentum states that the initial momentum should be equal to the final momentum, we can equate the two expressions for momentum:
Initial momentum = Final momentum
Momentum1 + Momentum2 = Momentum1' + Momentum2'
Now, let's substitute the expressions for momentum into this equation and solve for the unknowns v1 and v2.
For the 5.00 g object:
(mass1 * velocity1) + (mass2 * velocity2) = (mass1 * (-v1)) + (mass2 * (+v2))
Substituting the given values:
(5.00 g * 20.0 cm/s) + (10.0 g * 0 cm/s) = (5.00 g * (-v1)) + (10.0 g * (+v2))
Now, let's solve this equation to find the values of v1 and v2.
For (b), to find the fraction of the initial kinetic energy transferred to the 10.0 g object, we need to calculate the initial kinetic energy and the final kinetic energy of the 10.0 g object.
The initial kinetic energy of the system is given by:
Initial kinetic energy = (1/2) * mass1 * velocity1^2 + (1/2) * mass2 * velocity2^2
The final kinetic energy of the 10.0 g object is given by:
Final kinetic energy of 10.0 g object = (1/2) * mass2 * velocity2'^2
The fraction of the initial kinetic energy transferred to the 10.0 g object can be calculated as:
Fraction transferred = (Final kinetic energy of 10.0 g object) / (Initial kinetic energy) * 100
Now, let's substitute the given values of mass and velocities to calculate the velocities after the collision and the fraction of kinetic energy transferred.
momentum before = 5*20 = 100
momentum after = 100 = 5 v1 + 10 v2
Ke before = (1/2)5(400) = 1000
Ke after = 1000 = .5*5*v1^2 + .5*10*v2^2
or
2000 = 5 v1^2 + 10 v2^2
but v2 = (100-5v1)/10 = (10 -.5v1)
v2^2 = 100-10v1+.25v1^2
so
2000 = 5v1^1 + 10(100-10v1+.25v1^2)
solve quadratic for v1. That is it, go back and get v2 and the kinetic energies.
Now we did that all in cgs system rather than meters, kilograms, seconds. I assume you are using an old text. Your energies will be in dynes rather than joules for that reason.