On a frictionless, horizontal air table, puck A (with mass 0.25kg) is moving toward puck B (with mass 0.35kg), which is initially at rest.After the collision,puck A has a velocity of 0.12m/s to the left, and puck B has a velocity of 0.65m/s to the right.
a.what is speed of puck A before collision?
b.calculate the change in the total kinetic energyof the system that occurs during the collision
Let A's initial velocity be in the + direction (to the right)
If the two initial velocities are Va and Vb, then to conserve momentum we need
.25Va + .35*0 = .25(-0.12) + .35(0.65)
solve for Va
Now you can figure the total KE (1/2 mv^2) before and after
a. Well, it seems like puck A really wanted to show off its speed! Before the collision, the speed of puck A can be calculated using the conservation of momentum principle. The equation would be:
momentum before = momentum after
Since there are no external forces, the total momentum of the system is conserved. So we have:
(mass A × velocity A before) + (mass B × velocity B before) = (mass A × velocity A after) + (mass B × velocity B after)
Let's consider that puck B is initially at rest, so its velocity before the collision is 0. Substituting the given values:
(0.25 kg × velocity A before) + (0.35 kg × 0) = (0.25 kg × 0.12 m/s) + (0.35 kg × 0.65 m/s)
Simplifying the equation, you'll find the speed of puck A before the collision. But hey, I think puck A just wanted to keep its speed a mystery. Let's solve it and find out!
b. To calculate the change in total kinetic energy, we need the formula for kinetic energy:
KE = 0.5 × mass × velocity^2
In this case, the change in total kinetic energy would be:
ΔKE = KE after - KE before
Let's calculate it, but be careful, we don't want any kinetic energy to get lost during the process. It's like trying to catch a sneeze, you gotta be quick!
Remember, ΔKE can be positive or negative depending on whether the total kinetic energy increased or decreased. So hold your breath!
To answer your question, we'll use the principle of conservation of momentum.
a. To find the speed of puck A before the collision, we can determine the initial momentum of the system and divide it by the mass of puck A.
The momentum of an object is given by the formula: momentum = mass × velocity.
Before the collision, both pucks are separate entities, so the initial momentum is zero since puck B is initially at rest. After the collision, the momentum of the system is still zero because the total momentum before and after the collision must be conserved.
Therefore, we can set up the equation: momentum before collision = momentum after collision.
The momentum of puck A before the collision is:
momentum A(before collision) = mass A × velocity A (before collision)
momentum A = 0.25 kg × velocity A
Since we're trying to find the speed, we'll use the magnitude of the velocity. The magnitude of a velocity is always positive.
Therefore, we have:
magnitude of velocity A = magnitude of velocity A (before collision) = speed of A (before collision)
So, to find the speed of puck A before the collision, we can rewrite the equation as follows:
speed of A (before collision) = momentum A / mass A
Plugging in the values:
speed of A (before collision) = (0.25 kg × velocity A) / 0.25 kg
The mass of puck A cancels out:
speed of A (before collision) = velocity A
Therefore, the speed of puck A before the collision is the same as its velocity, which is 0.12 m/s to the left.
b. To calculate the change in the total kinetic energy during the collision, we can use the formula:
change in kinetic energy = final kinetic energy - initial kinetic energy
The kinetic energy of an object is given by the formula: kinetic energy = 0.5 × mass × velocity^2.
Before the collision, the total initial kinetic energy is the sum of the kinetic energies of the two pucks:
initial kinetic energy = 0.5 × mass A × velocity A^2 + 0.5 × mass B × velocity B^2
After the collision, the total final kinetic energy is the sum of the kinetic energies of the two pucks:
final kinetic energy = 0.5 × mass A × (velocity A (after collision))^2 + 0.5 × mass B × (velocity B (after collision))^2
To find the change in kinetic energy, we subtract the initial kinetic energy from the final kinetic energy:
change in kinetic energy = final kinetic energy - initial kinetic energy
Plugging in the values:
change in kinetic energy = [0.5 × 0.25 kg × (0.12 m/s)^2 + 0.5 × 0.35 kg × (0.65 m/s)^2] - [0.5 × 0.25 kg × (0 m/s)^2 + 0.5 × 0.35 kg × (0 m/s)^2]
Simplifying the equation:
change in kinetic energy = [0.5 × 0.25 kg × (0.12 m/s)^2 + 0.5 × 0.35 kg × (0.65 m/s)^2]
Therefore, the change in the total kinetic energy of the system during the collision is equal to the calculated value.
To find the speed of puck A before the collision, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision, assuming no external forces are acting on the system.
Let's denote the initial velocity of puck A as 'v_Ai' and the initial velocity of puck B as 'v_Bi'. The mass of puck A is 0.25 kg, so its momentum before the collision is given by m_A * v_Ai, where m_A = 0.25 kg. Since puck B is initially at rest, its momentum before the collision is 0.
After the collision, puck A has a velocity of 0.12 m/s to the left, which we can denote as '-0.12 m/s'. Puck B has a velocity of 0.65 m/s to the right, which is '+0.65 m/s'. The momentum of puck A after the collision is m_A * v_Af, where v_Af = -0.12 m/s. The momentum of puck B after the collision is m_B * v_Bf, where m_B = 0.35 kg and v_Bf = 0.65 m/s.
Using the conservation of momentum, we can write the equation:
m_A * v_Ai + m_B * v_Bi = m_A * v_Af + m_B * v_Bf
Plugging in the values, we get:
0.25 kg * v_Ai + 0.35 kg * 0 = 0.25 kg * (-0.12 m/s) + 0.35 kg * 0.65 m/s
Simplifying the equation:
0.25 kg * v_Ai = -0.03 kg m/s + 0.2275 kg m/s
Combining the terms on the right side:
0.25 kg * v_Ai = 0.1975 kg m/s
Now, we can solve the equation for v_Ai:
v_Ai = (0.1975 kg m/s) / 0.25 kg
v_Ai = 0.79 m/s
Therefore, the speed of puck A before the collision is 0.79 m/s.
To calculate the change in the total kinetic energy of the system during the collision, we need to find the initial kinetic energy and the final kinetic energy.
The initial kinetic energy of the system is given by:
KE_initial = (1/2) * m_A * v_Ai^2 + (1/2) * m_B * v_Bi^2
Substituting the values, we get:
KE_initial = (1/2) * 0.25 kg * (0.79 m/s)^2 + (1/2) * 0.35 kg * 0^2
Simplifying the equation:
KE_initial = (1/2) * 0.25 kg * (0.79 m/s)^2
KE_initial = 0.061425 J
The final kinetic energy of the system is given by:
KE_final = (1/2) * m_A * v_Af^2 + (1/2) * m_B * v_Bf^2
Substituting the values, we get:
KE_final = (1/2) * 0.25 kg * (-0.12 m/s)^2 + (1/2) * 0.35 kg * (0.65 m/s)^2
Simplifying the equation:
KE_final = (1/2) * 0.25 kg * (0.0144 m^2/s^2) + (1/2) * 0.35 kg * (0.4225 m^2/s^2)
KE_final = 0.0018 J + 0.0526125 J
KE_final = 0.0544125 J
The change in total kinetic energy during the collision is:
ΔKE = KE_final - KE_initial
ΔKE = 0.0544125 J - 0.061425 J
ΔKE = -0.0070125 J
Therefore, the change in the total kinetic energy of the system during the collision is -0.0070125 J (negative value implies a decrease in kinetic energy).
Given:
M1 = 0.25kg, V1 = ?.
M2 = 0.35 kg, V2 = 0.
V3 = -0.12 m/s = velocity of M1 after collision.
V4 = 0.65 m/s = velocity of M2 after collision.
Momentum before = Momentum after
a. M1*V1 + M2*V2 = M1*V3 + M2*V4.
0.25*V1 + 0.35*0 = 0.25*(-0.12) + 0.35*0.65,
0.25V1 = -0.03 + 0.228 - 0.35,
0.25V1 = -0.153,
V1 = -0.61 m/s = 0.61 m/s to the left.
b. Before collision:
KE = 0.5M1*V1^2 + 0.5M2*V2^2 = 0.5*0.25*(-0.61)^2 + 0 = 0.047 J.
After collision:
KE = 0.5M1*V3^2 + 0.5M2*V4^2.
KE = 0.5*0.25*(-0.12)^2 + 0.5*0.35*0.65^2 = ?
Change in KE = KE after - KE before.