Question

As shown in the figure below, object m1 = 1.60 kg starts at an initial height hij = 0.305 m and speed V1 = 4.00 m/s, swings downward and strikes in an elastic collision) object m2 = 4.55 kg which is initially at rest. 4.00 m/s m2 (a) Determine the speed of m(in m/s) just before the collision. X After you are convinced that energy is conserved as m, swings downward for the collision with me, see if you can write a statement of conservation of energy that will allow you to determine the speed of my as it collides with me. m/s (b) Determine the velocity (magnitude in m/s and direction) of each ball just after the collision (Assume the positive direction is toward the right. Indicate the direction with the sign of your answer.) mi 0 We have been informed that the collision is elastic and since no external forces are involved in the collision, momentum is also conserved. See if you can write a statement of conservation of momentum and a statement of conservation of energy which can be combined to obtain the final velocity of each mass. m/s x We have been informed that the collision is elastic and since no external forces are involved in the collision, momentum is also conserved. See if you can write a statement of conservation of momentum and a statement of conservation of energy which can be combined to obtain the final velocity of each mass. m/s m2 0 (c) Determine the height (in m) to which each ball swings after the collision (ignoring air resistance). m1 O Energy is conserved as m, and my swing upward to their maximum height. See if you can write a statement of conservation of energy that will allow you to determine the final height of me and my after the collision. m m2 X Energy is conserved as m, and m2 swing upward to their maximum height. See if you can write a statement of conservation of energy that will allow you to determine the final height of m, and my after the collision. m o

Answer 1

@ Mechanical Energy Conservation, 1 m, vrh + migh,i= 1 m, vith the +0 432 + 9,8% 0.305 2 2 V = 22 - V = 4.69 m/ Aus e=l= V2r - Vif 4.6940 24 - Vif a 4.69 – cij 1.6% 4.69 xl to= 1.6x Vif xl + 4.55 kfx l. = 7.5 1.64f + 4.55 from ear the cib f = -2.25m/ Ang Y = 2.94 m/s Ans. Mechanical energy conservation, & how zg he - 4 1 - 12.2004 - 10.26m Ause h = (9.94² 0*30 мм more ingles 19.6