two particles of mass m1=1.5 kg, and m2=3.5 kg undergo a 1-dimensional head-on collision. Their velocities along x are V1i=12 m/s, and V2i=-7.5 m/s. The 2 particles stick together after the collision. a.) Find the velocity after the collision. b.) How much kinetic energy is lost in the collision?
I used p for m1=18, and p for m2=-26.25, with a total system momentum of p=-8.25.
-8.25=1.5+3.5v
-8.25/5= v=-1.65 m/s.
Using KE=1/2mv^2, I calculated Kf to be -4.125, but then also calculated that Ki =-4.125.
Does this mean that there is no KE lost during the collision?
To determine if kinetic energy is lost in the collision, we need to consider the conservation of kinetic energy. In an elastic collision, where kinetic energy is conserved, the sum of the initial kinetic energies of the particles must be equal to the sum of the final kinetic energies.
Let's calculate the initial and final kinetic energies step-by-step:
a) Initial velocity of particle 1 (v1i) = 12 m/s
Mass of particle 1 (m1) = 1.5 kg
Initial kinetic energy of particle 1 (Ki1) = (1/2) * m1 * (v1i)^2
= (1/2) * 1.5 * (12)^2
= 108 J
b) Initial velocity of particle 2 (v2i) = -7.5 m/s
Mass of particle 2 (m2) = 3.5 kg
Initial kinetic energy of particle 2 (Ki2) = (1/2) * m2 * (v2i)^2
= (1/2) * 3.5 * (-7.5)^2
= 132.1875 J
c) Total initial kinetic energy (Ki) = Ki1 + Ki2
= 108 + 132.1875
= 240.1875 J
After the collision, the two particles stick together. Let's calculate the final velocity (vf) by using the conservation of momentum:
Total initial momentum = Total final momentum
m1 * v1i + m2 * v2i = (m1 + m2) * vf
1.5 * 12 + 3.5 * (-7.5) = (1.5 + 3.5) * vf
18 - 26.25 = 5 * vf
-8.25 = 5 * vf
vf = -8.25/5
vf = -1.65 m/s
d) Final velocity after the collision (vf) = -1.65 m/s
Now, let's calculate the final kinetic energy (Kf) using the final velocity:
Final kinetic energy (Kf) = (1/2) * (m1 + m2) * (vf)^2
= (1/2) * (1.5 + 3.5) * (-1.65)^2
= 4 * 1.65^2 / 2
= 10.89 J
e) Kinetic energy lost in the collision = Ki - Kf
= 240.1875 - 10.89
= 229.2975 J
Therefore, the amount of kinetic energy lost in the collision is approximately 229.2975 J.
To solve this problem, we can use the principle of conservation of momentum and the principle of conservation of kinetic energy.
a) Conservation of momentum:
The initial momentum of the system is given by the sum of the individual momenta of the two particles:
p_initial = m1 * V1i + m2 * V2i
= (1.5 kg)(12 m/s) + (3.5 kg)(-7.5 m/s)
= 18 kg*m/s - 26.25 kg*m/s
= -8.25 kg*m/s
After the collision, the two particles stick together and move with a combined velocity (Vf). Applying the conservation of momentum:
p_final = (m1 + m2) * Vf
= (1.5 kg + 3.5 kg)(Vf)
= 5 kg * Vf
Setting the initial and final momenta equal to each other:
p_initial = p_final
-8.25 kg*m/s = 5 kg * Vf
Solving for Vf:
Vf = -8.25 kg*m/s / 5 kg
= -1.65 m/s
So the velocity after the collision is -1.65 m/s.
b) Conservation of kinetic energy:
The initial kinetic energy of the system is given by the sum of the individual kinetic energies of the two particles:
KE_initial = 1/2 * m1 * (V1i)^2 + 1/2 * m2 * (V2i)^2
= 1/2 * (1.5 kg)(12 m/s)^2 + 1/2 * (3.5 kg)(-7.5 m/s)^2
= 108 J + 196.875 J
= 304.875 J
The final kinetic energy of the combined particles is given by:
KE_final = 1/2 * (m1 + m2) * (Vf)^2
= 1/2 * (1.5 kg + 3.5 kg) * (-1.65 m/s)^2
= 2.5 kg * 2.7225 m^2/s^2
= 6.80625 J
The change in kinetic energy is given by:
ΔKE = KE_final - KE_initial
= 6.80625 J - 304.875 J
= -298.06875 J
So, the kinetic energy lost in the collision is -298.06875 J.
Now to address your calculations:
It seems that there was an error in your calculation of the final velocity. The correct value is -1.65 m/s. Regarding the kinetic energy, it can indeed be confusing because you obtained the same value for the initial and final kinetic energy (-4.125 J). However, this does not mean that no kinetic energy was lost during the collision. The negative sign indicates a decrease in kinetic energy. In this case, the change in kinetic energy (-298.06875 J) represents the amount of energy lost during the collision.