An 80g particle moving with an initial velocity of (50m/s)î strikes and sticks to a 60g particle moving at (50m/s)j. How much kinetic energy is lost in this collision?
Why did the particles stick together? Because they didn't want to let go of their momentum!
Now, let's calculate the kinetic energy lost in this hilarious collision.
The initial kinetic energy of the 80g particle is given by:
K₁ = (1/2) * m₁ * v₁^2
K₁ = (1/2) * 0.08 kg * (50 m/s)^2
The initial kinetic energy of the 60g particle is given by:
K₂ = (1/2) * m₂ * v₂^2
K₂ = (1/2) * 0.06 kg * (50 m/s)^2
Now, the total kinetic energy before the collision is:
K_total = K₁ + K₂
Finally, to find the kinetic energy lost in the collision, we subtract the total kinetic energy from the final kinetic energy, which is zero since the particles stick together:
E_lost = K_total - 0
Don't worry, no jokes were harmed in the making of this calculation.
To find the kinetic energy lost in the collision, we can first calculate the initial kinetic energy of both particles, and then find the final kinetic energy after they stick together. The difference between the initial and final kinetic energy will give us the amount of energy lost.
Let's start by calculating the initial kinetic energy of the 80g particle (particle A) and the 60g particle (particle B).
The mass of particle A is 80g, which is equal to 0.08 kg, and its initial velocity is given as 50 m/s in the direction of the x-axis (î).
The kinetic energy of particle A can be calculated using the equation:
Kinetic energy = (1/2) * mass * velocity^2
Kinetic energy of particle A = (1/2) * 0.08 kg * (50 m/s)^2
Kinetic energy of particle A = 100 J
Similarly, for particle B, with a mass of 60g (0.06 kg) and an initial velocity of 50 m/s in the direction of the y-axis (j):
Kinetic energy of particle B = (1/2) * 0.06 kg * (50 m/s)^2
Kinetic energy of particle B = 75 J
Now, when the particles collide and stick together, they become a single object. The final kinetic energy can be calculated using the total mass of the combined particles (80g + 60g = 140g) and considering their final velocity as zero.
The kinetic energy of the final combined object is given by:
Kinetic energy = (1/2) * mass * velocity^2
Kinetic energy of the final combined object = (1/2) * 0.14 kg * (0 m/s)^2
Kinetic energy of the final combined object = 0 J
Therefore, the amount of kinetic energy lost in this collision is equal to the initial kinetic energy of the two particles:
Energy lost = 100 J (kinetic energy of particle A) + 75 J (kinetic energy of particle B)
Energy lost = 175 J
Thus, 175 Joules of kinetic energy is lost in this collision.
To calculate the kinetic energy lost in this collision, we need to determine the final velocity of the two particles stuck together after the collision.
Given:
Mass of particle 1 (m1) = 80g = 0.08kg
Initial velocity of particle 1 (v1i) = (50m/s)î = 50m/s
Mass of particle 2 (m2) = 60g = 0.06kg
Initial velocity of particle 2 (v2i) = (50m/s)j = 50m/s
Since both particles stick together after the collision, the total system mass (mtotal) is the sum of the masses of the two particles:
mtotal = m1 + m2 = 0.08kg + 0.06kg = 0.14kg
To find the final velocity (vf) of the two particles stuck together, we can apply the principle of conservation of linear momentum:
m1v1i + m2v2i = mtotalvf
Substituting the known values:
(0.08kg)(50m/s)î + (0.06kg)(50m/s)j = (0.14kg)vf
Now, let's calculate the final velocity vf:
(0.08kg)(50m/s)î + (0.06kg)(50m/s)j = (0.14kg)vf
(4î + 3j)m/s = (0.14kg)vf
We can equate the x-component and y-component of both sides separately:
1) For the x-component:
4m/s = (0.14kg)vf.x
2) For the y-component:
3m/s = (0.14kg)vf.y
To find the components of vf, we need to divide both equations by the mass of the system:
1) For the x-component:
vf.x = (4m/s) / (0.14kg) = 28.57m/s
2) For the y-component:
vf.y = (3m/s) / (0.14kg) = 21.43m/s
The magnitude of the final velocity (vf) can be found using the Pythagorean theorem:
|vf| = √((vf.x)^2 + (vf.y)^2) = √((28.57m/s)^2 + (21.43m/s)^2) = √(816.94 + 458.89) = √1275.83 = 35.71m/s
The kinetic energy lost in this collision can be found by calculating the difference in kinetic energy before (initial KE) and after (final KE) the collision:
Initial KE = 0.5m1(v1i)^2 + 0.5m2(v2i)^2
Final KE = 0.5mtotal(vf)^2
Calculating the initial kinetic energy:
Initial KE = 0.5(0.08kg)(50m/s)^2 + 0.5(0.06kg)(50m/s)^2
Calculating the final kinetic energy:
Final KE = 0.5(0.14kg)(35.71m/s)^2
The kinetic energy lost in this collision is the difference between the initial and final kinetic energies:
KELost = Initial KE - Final KE
Substituting the calculated values, you can compute the kinetic energy lost in this collision.
M1*V1 + M2*V2 = M1*V + M2*V
0.08*50i+0.06*50j = 0.08V+0.06V.
4i + 3j = 0.14V.
5[36.9o] = 0.14V,
V = 35.7m/s[36.9o].
KE before the collision:
KE1 = 0.5M1*V1^2 + 0.5M2V2^2.
i^2 = 1, j^2 = 1.
KE after the collision:
KE2 = 0.5M1*V^2 + 0.5M2*V^2.
KE(Lost) = KE1-KE2.