Elastic collision in 2D?
On a frictionless surface, a 0.35 kg puck moves horizontally to the right (at an angle of 0°) and a speed of 2.3 m/s. It collides with a 0.23 kg puck that is stationary. After the collision, the puck that was initially moving has a speed of 2.0 m/s and is moving at an angle of −32°. What is the velocity of the other puck after the collision in m/s and at what angle in °?
- Anonymous1 decade agoFavorite Answer
This is just a conservation of momentum question.
The initial momentum of the system must equal the final momentum of the system.
Momentum (p) = mass (m) multiplied by velocity (v),
p = mv
In the initial case, only the fist puck is moving and has all of its momentum pointed along the +X axis (0° direction).
In the final case, both pucks are moving and have both an X and Y component to their momentum. But the TOTAL momentum of both pucks put together still equals the initial momentum of the first puck.
In solving, we will consider each component of the momentum (X and Y) separately since the perpendicular vectors are independent.
We will say that _x indicates the X component of the quantity being discussed, and similarly, _y indicates the Y component of the quantity being discussed.
We will say that _1 and _2 refers to the first and second puck respectively and that _i and _f indicated the initial and final quantities of the quantity (i.e. before and after the collision).
P_x_i = m_1_i * v_1_i * cos (theta_1_i)
P_y_i = m_1_i * v_1_i * sin (theta_1_i)
Since theta = 0°, cos(theta_1_i) = 1, sin (theta_1_i) = 0, and,
P_x_i = m_1 * v_1_i = (.35 kg) * (2.3 m/s) = .805 kg m/s
P_y_i = 0 kg m/s
P_x_f = m_1 * v_1_f * cos (theta_1_f) + m_2 * v_2_f * cos (theta_2_f)
P_y_f = m_1 * v_1_f * sin (theta_1_f) + m_2 * v_2_f * sin (theta_2_f)
We are not given a value for theta_2_f, but we are told that theta_1_f = -32°. Plugging in, we get,
P_x_f = m_1 * v_1_f * cos (-32°) + m_2 * v_2_f * cos (theta_2_f)
P_y_f = m_1 * v_1_f * sin (-32°) + m_2 * v_2_f * sin (theta_2_f)
Plugging in some more,
P_x_f = (.35 kg) * (2 m/s) * cos (-32°) + (.23 kg) * v_2 * cos (theta_2_f)
P_y_f = (.35 kg) * (2 m/s) * sin (-32°) + (.23 kg) * v_2 * sin (theta_2_f)
Evaluating where convenient,
P_x_f = (.594 kg m/s) + (.23 kg) * v_2_f * cos (theta_2_f)
P_y_f = (-.371 kg m/s) + (.23 kg) * v_2_f * sin (theta_2_f)
We know that p_x_f = p_x_i, and p_y_f = p_y_i, setting these two equal we get,
p_x_f = p_x_i
(.594 kg m/s) + (.23 kg) * v_2_f * cos (theta_2_f) = .805 kg m/s
Which we can solve for v_2_f * cos (theta_2_f) as,
v_2_f * cos (theta_2_f) = .917 m/s
p_y_f = p_y_i
(-.371 kg m/s) + (.23 kg) * v_2_f * sin (theta_2_f) = 0 kg m/s
So now we can solve for v_2_f * sin (theta_2_f) as,
v_2_f * sin (theta_2_f) = 1.613 m/s
We now know the final X and Y components of the 2nd pucks velocity. Combining these vectors we can get,
v _2_f = sqrt ((v _2_f_x)^2 + (v _2_f_y)^2)
v _2_f = sqrt (.917^2 + 1.613^2)
v _2_f = 1.855 m/s
This is the final speed of the 2nd puck after the collision…but now to find its velocity (speed and direction).
We know that v_2_f_y = v_2_f_y * sin (theta_2_f)
The only thing we don’t know is the angle, theta_2_f, but we can now solve for it.
sin (theta_2_f) = v_2_f_y / v_2_f_y
theta_2_f = arcsin (v_2_f_y / v_2_f_y) = arcsin ((1.613 m/s) / (1.855 m/s))
theta_2_f = 60.4°.
Similarly, we can arrive at the same answer by using the X velocity,
cos (theta_2_f) = v_2_f_y / v_2_f_y
theta_2_f = arccos (v_2_f_x / v_2_f_x) = arccos ((.917 m/s) / (1.855 m/s))
theta_2_f = 60.4°.
The numbers match (…which is a good sign).
So the final velocity of the second puck after the collision is 1.86 m/s at an angle of 60.4° above the horizontal.
- leitchLv 44 years ago
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