A 0.300 kg puck, initially at rest on a frictionless horizontal surface, is stru
ID: 1960666 • Letter: A
Question
A 0.300 kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.205 kg puck that is initially moving along the x axis with a velocity of 2.15 m/s. After the collision, the 0.205 kg puck has a speed of 1.10 m/s at an angle of q = 56.3o to the positive x axis. Determine the velocity of the 0.300 kg puck after the collsion.Calculate the magnitude of the angle with respect to the x-axis.
Find the percentage of kinetic energy lost in the collision.
Im sorry this is such a long question but i have tried it multiple times using the class notes and i cant seem to get the correct answer and i am down to my last few attempts. Thank you!
Explanation / Answer
First, use conservation of momentum to find the velocity of the other puck. Using capital letters for vectors (and _ for subscipts): m_1*V_1 + m_2*V_1 = m_1*V_1' + m_2*V_2' m_1 = 0.3 kg = mass of puck 1 m_2 = 0.205 = mass of puck 2 V_1 = (v_1x, v_1y) = (0, 0) = initial velocity of puck 1 V_2 = (v_2x, v_2y) = (2.15, 0) m/s = initial velocty of puck 2 V_1' and V_2' are the final velocities of the pucks. You have: ||V_2'|| = 1.10 m/s with a direction of ? = 56.3° with respect to +x. v_2x' = ||V_2|| cos ? v_2y' = ||V_2|| sin ? So, the x component of momentum solves for v1_x' by: m1*v_1x + m2*v_2x = m1*v_1x' + m2*v2_x' v1_x' = (m1*v_1x + m2*v_2x - m2*v2_x') / m1 Similarly, v1_y' = (m1*v_1y + m2*v_2y - m2*v_2y') / m1 That's V_2' in (x,y) terms. Compute ||V_2'|| using Pythagoras and then angle by: ?_1' = arctan(v_2y' / v_2x') That's V_2' in magnitude and direction form. For kinetic energy loss, compute the kinetic energies before: K_1 = 0 K_2 = (1/2) m2 * ||V_2||² K = K_1 + K_2 ...and after: K_1' = (1/2) m1 * ||V_1||² K_2' = (1/2) m2 * ||V_2||² K' = K_1' + K_2' % loss = 100% * (K - K')/K So, there are a lot of pieces, but each step is straightforward. Get used to converting to Cartesian form (x,y) for vector addition problems. It makes things much simpler. In this case, you solve independently for x and y final momentum with a simple conservation formula.
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