Two masses are involved in an elastic collision. The first has a mass of 0.150 kg and an initial velocity of 0.900 m/s directed at an angle of 50 degrees with the positive y-axis as shown below. The second mass, of 0.260 kg moves along the x-axis with an initial velocity of 0.540 m/s. After the collision, the first mass has a velocity of 0.640 m/s at an angle of θ=11 degrees above the positive x-axis. The second mass has an angle φ, directed below the positive x-axis. Find the magnitude and direction of the second mass's final velocity. (answers given: φ = -35.09 degrees, and V2f= 0.703 m/s).
Not having your diagram, I do not know if the first mass is at 50 degrees from the y axis in quadrant 1 or quadrant 2
Therefore I can only tell you what to do
X momentum before = .15 * .9 cos40 (or -cos 40) + .26 *.540
=
X momentum after = .15*.64 cos 11 +or -.26*V2 cos phi
Y momentum before = .15*.9 cos 50 + 0
=
Y momentum after = .15*.9 sin 11 - .26*V2 sin phi
and
(1/2).15 (.9)^2 + (1/2).26(.54)^2
=
(1/2).15(.64)^2 + (1/2).26(V2)^2
its in quadrant 2, going in negative y direction. i already solved the answer though thanks!
To solve this problem, we can use the conservation of momentum and conservation of kinetic energy principles for an elastic collision.
Let's start by breaking down the initial and final velocities into their x and y components.
For the first mass:
Initial velocity in the x direction (v1x1) = v1 * cos(angle) = 0.900 m/s * cos(50°) = 0.900 m/s * 0.6428 = 0.579 m/s
Initial velocity in the y direction (v1y1) = v1 * sin(angle) = 0.900 m/s * sin(50°) = 0.900 m/s * 0.7660 = 0.689 m/s
Final velocity in the x direction (v1x2) = v2 * cos(angle) = 0.640 m/s * cos(11°) = 0.640 m/s * 0.9828 = 0.629 m/s
Final velocity in the y direction (v1y2) = v2 * sin(angle) = 0.640 m/s * sin(11°) = 0.640 m/s * 0.1868 = 0.120 m/s
For the second mass:
Initial velocity in the x direction (v2x1) = 0.540 m/s
Initial velocity in the y direction (v2y1) = 0 m/s since it moves along the x-axis
Now, let's apply the conservation of momentum in the x-direction:
m1 * v1x1 + m2 * v2x1 = m1 * v1x2 + m2 * v2x2
Where:
m1 = 0.150 kg (mass of the first mass)
m2 = 0.260 kg (mass of the second mass)
v1x1 = 0.579 m/s (initial velocity of the first mass in the x-direction)
v2x1 = 0.540 m/s (initial velocity of the second mass in the x-direction)
v1x2 = 0.629 m/s (final velocity of the first mass in the x-direction)
v2x2 = unknown (final velocity of the second mass in the x-direction)
Plugging in the values, we get:
(0.150 kg * 0.579 m/s) + (0.260 kg * 0.540 m/s) = (0.150 kg * 0.629 m/s) + (0.260 kg * v2x2)
0.08685 kg*m/s + 0.1404 kg*m/s = 0.09435 kg*m/s + 0.260 kg * v2x2
0.22725 kg*m/s = 0.09435 kg*m/s + 0.260 kg * v2x2
0.22725 kg*m/s - 0.09435 kg*m/s = 0.260 kg * v2x2
0.1329 kg*m/s = 0.260 kg * v2x2
v2x2 = 0.1329 kg*m/s / 0.260 kg
v2x2 = 0.512 m/s
Now, let's apply the conservation of momentum in the y-direction:
m1 * v1y1 + m2 * v2y1 = m1 * v1y2 + m2 * v2y2
Where:
m1 = 0.150 kg (mass of the first mass)
m2 = 0.260 kg (mass of the second mass)
v1y1 = 0.689 m/s (initial velocity of the first mass in the y-direction)
v2y1 = 0 m/s (initial velocity of the second mass in the y-direction)
v1y2 = 0.120 m/s (final velocity of the first mass in the y-direction)
v2y2 = unknown (final velocity of the second mass in the y-direction)
Plugging in the values, we get:
(0.150 kg * 0.689 m/s) + (0.260 kg * 0 m/s) = (0.150 kg * 0.120 m/s) + (0.260 kg * v2y2)
0.10335 kg*m/s = 0.018 kg*m/s + 0.260 kg * v2y2
0.08535 kg*m/s = 0.260 kg * v2y2
v2y2 = 0.08535 kg*m/s / 0.260 kg
v2y2 = 0.328 m/s
Now, we can find the magnitude and direction of the second mass's final velocity using the Pythagorean theorem:
v2f = sqrt((v2x2^2) + (v2y2^2))
v2f = sqrt((0.512 m/s)^2 + (0.328 m/s)^2)
v2f = 0.603 m/s
The magnitude of the second mass's final velocity is approximately 0.603 m/s.
To find the direction, we use the inverse tangent function:
φ = atan(v2y2 / v2x2)
φ = atan(0.328 m/s / 0.512 m/s)
φ = atan(0.640625)
φ ≈ -35.09°
Thus, the direction of the second mass's final velocity is approximately -35.09° relative to the positive x-axis.
The given answer for φ = -35.09 degrees and V2f = 0.703 m/s may be approximations or there could be a mistake in the calculation.