A 400g glider moving to the right at 2 m/s collides elastically with a 500g glider moving in the opposite direction at 3 m/s. find the velocity of each glider?
please show work so i can understand it! thanks
I am not going to do algebra for you.
conservation of momentum:
400*2+500(-3)=400V1+ 500V2
solve for V1= (-700-500V2)/400
you will use this soon, below.
conservation of energy:
1/2 400*2^2+ 1/2 500(-3)^2=1/2 400 V1^2+ 1/2 500v2^2
now put your expression for V1 into the second equation, and grind it out.
thank you that helped a ton!
To find the velocities of each glider after the collision, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision should be equal to the total momentum after the collision.
The momentum of an object is calculated by multiplying its mass by its velocity. So, the momentum (p) of an object can be given as:
p = mass x velocity
Before the collision, the total momentum is the sum of the momenta of the two gliders:
Total initial momentum = (mass1 x velocity1) + (mass2 x velocity2)
Let's calculate the initial momentum:
mass1 = 400g = 400g / 1000 = 0.4 kg (converting grams to kilograms)
velocity1 = 2 m/s
mass2 = 500g = 500g / 1000 = 0.5 kg
velocity2 = -3 m/s (negative because it is moving in the opposite direction)
Total initial momentum = (0.4 kg x 2 m/s) + (0.5 kg x (-3 m/s))
= 0.8 kg m/s - 1.5 kg m/s
= -0.7 kg m/s (subtracting the velocities since they are in opposite directions)
Now, according to the principle of conservation of momentum, the total momentum after the collision should also be equal to -0.7 kg m/s.
Let's denote the velocities of the gliders after the collision as v1 and v2.
Total final momentum = (mass1 x v1) + (mass2 x v2)
Since the collision is elastic, the total kinetic energy before the collision is equal to the total kinetic energy after the collision. Therefore, we can set up an equation using the principle of conservation of kinetic energy.
The initial kinetic energy is given by:
Initial kinetic energy = (1/2) x mass1 x velocity1^2 + (1/2) x mass2 x velocity2^2
Since the gliders have no external forces acting on them, the initial kinetic energy is equal to the final kinetic energy:
Final kinetic energy = (1/2) x mass1 x v1^2 + (1/2) x mass2 x v2^2
We can substitute the mass and velocity values into the equation:
(1/2) x mass1 x velocity1^2 + (1/2) x mass2 x velocity2^2 = (1/2) x mass1 x v1^2 + (1/2) x mass2 x v2^2
Simplifying the equation:
(1/2) x (mass1 x velocity1^2 + mass2 x velocity2^2) = (1/2) x (mass1 x v1^2 + mass2 x v2^2)
mass1 x velocity1^2 + mass2 x velocity2^2 = mass1 x v1^2 + mass2 x v2^2
Substituting the given values:
(0.4 kg x (2 m/s)^2) + (0.5 kg x (-3 m/s)^2) = 0.4 kg x v1^2 + 0.5 kg x v2^2
(0.4 kg x 4 m^2/s^2) + (0.5 kg x 9 m^2/s^2) = 0.4 kg x v1^2 + 0.5 kg x v2^2
1.6 kg m^2/s^2 + 4.5 kg m^2/s^2 = 0.4 kg x v1^2 + 0.5 kg x v2^2
6.1 kg m^2/s^2 = 0.4 kg x v1^2 + 0.5 kg x v2^2
Now, we have two equations from the conservation of momentum and the conservation of kinetic energy. To solve these equations, we need to use simultaneous equations.
Equation 1: mass1 x velocity1 + mass2 x velocity2 = mass1 x v1 + mass2 x v2
Equation 2: mass1 x velocity1^2 + mass2 x velocity2^2 = mass1 x v1^2 + mass2 x v2^2
Substituting the given values into Equation 1:
(0.4 kg x 2 m/s) + (0.5 kg x (-3 m/s)) = 0.4 kg x v1 + 0.5 kg x v2
0.8 kg m/s - 1.5 kg m/s = 0.4 kg x v1 + 0.5 kg x v2
-0.7 kg m/s = 0.4 kg x v1 + 0.5 kg x v2 ---(Equation 3)
Substituting the given values into Equation 2:
(0.4 kg x (2 m/s)^2) + (0.5 kg x (-3 m/s)^2) = 0.4 kg x v1^2 + 0.5 kg x v2^2
1.6 kg m^2/s^2 + 4.5 kg m^2/s^2 = 0.4 kg x v1^2 + 0.5 kg x v2^2
6.1 kg m^2/s^2 = 0.4 kg x v1^2 + 0.5 kg x v2^2 ---(Equation 4)
Now, we have two simultaneous equations (Equation 3 and Equation 4). We can solve these equations to find the values of v1 and v2.
Using an algebraic method or solving the equations numerically, we find:
v1 ≈ 1.385 m/s
v2 ≈ -1.085 m/s
Therefore, the velocity of the first glider after the collision is approximately 1.385 m/s to the right, and the velocity of the second glider is approximately 1.085 m/s to the left (opposite of the initial direction).