Question

As shown in the figure, a wooden ball with mass m2 is initially at rest on a horizontal, frictionless table. A second- en mass m, moving with a speed 2.00 m/s, collides with m2. Assume m, moves initially along the +x-axis. After the collision, m moves with speed 1.00 m/s at an angle of θ-53.0° to the positive x-axis. (Assume m1-0.200 kg and m2-0.300 kg.) After the collision Vif / sin θ ycos 8 Before the collision li (a) Determine the speed (in m/s) of the 0.300 kg ball after the collision. m/s (b) Find the fraction of kinetic energy transferred away or transformed to other forms of energy in the colsion IAKİ Note that the

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Answer 1

Apply conservation of momentum, m_1 v_1 i = m_i v_1 F cos theta + m_2 v_2 F cos phi 0.2 times 2 = 0.2 times 2 = 0.2 times 1 times cos 53 degree + 0.3 V_2 + cos phi 0.3 V_2 F cos phi = 0.279636995 V_2 F cos phi = 0.932123317 1 m_1 v_1 F sin theta = m_2 v_2 F sin phi V_2 + sin phi = 0.532423673 2 from 1 & 2 V_2 F = squareroot (0.932123317)^2 + (0.532423673)^2 = 1.07 m/sVertical-bar Delta K Vertical-bar Ki = 1/2 m_1 v_1i^2 - (1/2 m_1 v_1 F^2 + 1/2 m_2 v^2_2f)/1/2 m_1 v^2_1i \r\n Delta k/ki = 1 - (1/2 times 0.2 times 1^2 + 1/2 times 0.3 times 1.07^2)/1/1/2 times 0.2 times 2^2Vertical-bar Delta k Vertical-bar /k = 0.32