A hockey puck (mass mA = 0.10kg) moves along the ice in the x-direction at a spe

Question

A hockey puck (mass mA = 0.10kg) moves along the ice in the x-direction at a speed of 8.0m/s. It strikes a second puck (mB = 0.40kg) which is initially at rest. After the collision, it is observed that puck A moves off at an angle of ? below the x-axis. a/ Write the equations taht represent conservatin of momentum (2 components: x and y) for this event, including known values. b/ If the final angles are measured to have the values ?=60.00 and f=30.00, determine the final speeds for the two pucks. Calculate the initial and final translational kinetic energy for the system. A hockey puck (mass mA = 0.10kg) moves along the ice in the x-direction at a speed of 8.0m/s. It strikes a second puck (mB = 0.40kg) which is initially at rest. After the collision, it is observed that puck A moves off at an angle of ? below the x-axis. a/ Write the equations taht represent conservatin of momentum (2 components: x and y) for this event, including known values. b/ If the final angles are measured to have the values ?=60.00 and f=30.00, determine the final speeds for the two pucks. Calculate the initial and final translational kinetic energy for the system.

Explanation / Answer

momentum is conserved also energy should be (we'll see farther down) momentum is a vector quantity initial momentum is M1V1 because that's all that's moving the momentum is the same total after the collision M1V1?0 deg = M1V2?? + M2V3?? I'm not sure why we want the sum as x,y but M1V1?0 = (M1V1, 0) M1V2?? = (M1V2*cos?, M1V2*sin?) M2V3?? = (M2V3*cos?, M2V3*sin?) or (M1V1, 0) = (M1V2*cos?, M1V2*sin?) + (M2V3*cos?, M2V3*sin?) the x components add to M1V1 and y components add to zero (Y1 = -Y2) if we draw the diagram with 60 deg and 30 deg we get a right triangle so (M1V1)^2 = (M1V2)^2 + (M2V3)^2 (0.10 * 8)^2 = (0.10V2)^2 + (0.40V3)^2 going back to (Y1 = -Y2) M1V2*sin? = -M2V3*sin? 0.10*V2*0.5 = -040*V3*(-0.866) (note that ? is -60 deg since we let ? be +30 deg) V2 = 6.928 V3 substitute into (0.10 * 8)^2 = (0.10V2)^2 + (0.40V3)^2 and divide through by 0.10^2 gives 64 = 48 V3^2 + 16 V3^2 V3 = 1.00 m/s V2 = 6.928 V3= 6.93 m/s the initial KE = 1/2 M1V1^2 = 1/2*0.10*8^2 = 3.2 J final is 1/2M1V2^2 + 1/2 M2V3^2 = 1/2*0.10*(6.93)^2 + 1/2*0.40*1^2 = 2.40 + 0.2 = 2.6 J

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