An air track glider of mass m1 = 0.250 kg moving at 0.900 m/s to the right collides with a glider of mass m2 = 0.500 kg at rest. If m1 rebounds and moves to the left with a speed of 0.300 m/s, what is the speed and direction of m2 after the collision? kinetic energy

The speed of m2 is 0.6 m/s and its direction is to the right.

Explanation:

This numerical can be solved easily by applying law of conservation of momentum to it. According to law of conservation of momentum:

Total Momentum Before Collision = Total Momentum After Collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where,

m₁ = Mass of 1st air glider = 0.25 kg

m₂ = Mass of 2nd air glider = 0.5 kg

u₁ =  Speed of 1st air glider before collision = 0.9 m/s

u₂ = Speed of 2nd air glider before collision = 0 m/s (at rest)

v₁ =  Speed of 1st air glider after collision = - 0.3 m/s (negative sign due to change in direction of velocity)

v₂ = Speed of 2nd air glider after collision = ?

Therefore,

(0.25 kg)(0.9 m/s) + (0.5 kg)(0 m/s) = (0.25 kg)(-0.3 m/s) + (0.5 kg)v₂

0.225 kg.m/s + 0.075 kg.m/s = (0.5 kg)v₂

v₂ = (0.3 kg.m/s)/(0.5 kg)

v₂ = 0.6 m/s

Positive sign indicates that v₂ is directed towards right

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14