Three particles of equal masses travelling with velocities of 10m/s,20m/s and 30m/s respectively along x-axis at an angle of 30° to the direction of positive x-axis and Y-axis collide simultaneously and get sticked to each other
1)The combined particle will move with velocity
Ans:10/3√20+2√3m/s
Combined particle will move at angle α wth X-axis where
Ans: α=taninverse [2(√3+2)]
1. M*V1 + M*V2 + M*V3 = M*V + M*V + M*V.
M*10[30o] + M*20[30] + M*30[30] = 3MV.
Divide both sides by M:
10[30o] + 20[30] + 30[30] = 3V,
60[30o] = 3V,
V = 20m/s[30o].
Pls send the solution how like that answers will get in the above question
10/3(√20+2√3)
To solve this problem, we can use the principle of conservation of momentum. The total momentum before the collision should be equal to the total momentum after the collision.
1) Calculating the initial momentum:
The momentum of a particle is given by the product of its mass and velocity.
Let's denote the masses of the three particles as m.
Particle 1: 10 m/s
Particle 2: 20 m/s
Particle 3: 30 m/s
The components of the velocities along the x-axis can be found using trigonometry.
Particle 1: vx1 = 10 m/s * cos(30°)
Particle 2: vx2 = 20 m/s * cos(30°)
Particle 3: vx3 = 30 m/s * cos(30°)
The components of the velocities along the y-axis can be found using trigonometry.
Particle 1: vy1 = 10 m/s * sin(30°)
Particle 2: vy2 = 20 m/s * sin(30°)
Particle 3: vy3 = 30 m/s * sin(30°)
The total momentum before the collision along the x-axis is given by:
P_initial_x = m * (vx1 + vx2 + vx3)
The total momentum before the collision along the y-axis is given by:
P_initial_y = m * (vy1 + vy2 + vy3)
2) Calculating the final momentum:
After the collision, the three particles stick together and move as a single combined particle.
The velocity of the combined particle can be calculated as the center of mass velocity of all particles.
v_combined_x = (vx1 + vx2 + vx3)/3
v_combined_y = (vy1 + vy2 + vy3)/3
The total momentum after the collision along the x-axis is given by:
P_final_x = m * v_combined_x
The total momentum after the collision along the y-axis is given by:
P_final_y = m * v_combined_y
3) Applying conservation of momentum:
According to the principle of conservation of momentum, the total initial momentum should be equal to the total final momentum.
P_initial_x = P_final_x
P_initial_y = P_final_y
m * (vx1 + vx2 + vx3) = m * v_combined_x
m * (vy1 + vy2 + vy3) = m * v_combined_y
Canceling out the mass m from both sides, we get:
(vx1 + vx2 + vx3) = v_combined_x
(vy1 + vy2 + vy3) = v_combined_y
4) Calculating the combined particle's velocity and angle:
Now we can substitute the values of vx1, vx2, vx3, vy1, vy2, and vy3 into equations (1) and (2) above to find the values of v_combined_x and v_combined_y.
v_combined_x = (10 m/s * cos(30°) + 20 m/s * cos(30°) + 30 m/s * cos(30°))/3
v_combined_y = (10 m/s * sin(30°) + 20 m/s * sin(30°) + 30 m/s * sin(30°))/3
Simplifying further, we get:
v_combined_x = (10 + 20 + 30)/3 * cos(30°)
v_combined_y = (10 + 20 + 30)/3 * sin(30°)
v_combined_x = 60/3 * cos(30°)
v_combined_y = 60/3 * sin(30°)
v_combined_x = 20 * cos(30°)
v_combined_y = 20 * sin(30°)
Using trigonometric identities, we can simplify further:
v_combined_x = 20 * √3/2
v_combined_y = 20 * 1/2
v_combined_x = 10√3
v_combined_y = 10
Finally, we can find the magnitude and angle of the combined particle's velocity using the Pythagorean theorem and the inverse tangent function:
v_combined = √(v_combined_x^2 + v_combined_y^2)
α = tan^(-1)(v_combined_y/v_combined_x)
Substituting the values, we get:
v_combined = √((10√3)^2 + 10^2) = √(300 + 100) = √400 = 20 m/s
α = tan^(-1)(10/10√3) = tan^(-1)(1/√3) = 30°
Therefore, the combined particle will move with a velocity of 20 m/s at an angle of 30° with the positive x-axis.
So, the given answer is incorrect.
v_combined = 20 m/s
α = 30°