This print-out should have 9 questions. Multiple choice questions may continue o

Question

This print-out should have 9 questions. Multiple choice questions may continue on the next column or page - find all choices before answering. A(n) 7 kg object moving with a speed of to the right, toward the 5 kg block. The 9.2 m/s collides with a(n) 19 kg object moving surface is with a velocity of 11 m/s in a direction 22 degree from the What is the speed of the two objects after together) just after they collide. the collision if they remain stuck together? Answer in units of m/Answer in units of m 006 (part 4 of 4) 10.0 points Find the horizontal distance the blocks move 002 (part 2 of 2) 10.0 point what is the change in direction experienced before coming to rest. by the lighter of the two objects? Answer in units of m. Answer in units of 007 10.0 points A 11.6g bullet is fired vertically into a 5.06 kg 003 (part 1 of 4) 10.0 points A mass less spring with force constant block of wood. 510 N/m is fastened at its left end to a vertical wall, as shown below. The acceleration of gravity is 9.8 m/s 510 N/m. The bullet gets stuck in the block, and the 5 kg pact lifts the block 0.013 m up. (That is, the block-with the bullet stuck in it-rises initially, the 7 kg block and 5 kg block rest 0.013 m up above its initial position, and then on a horizontal surface with the 7 kg block in falls back down.) M/s What was the in Given g contact with the spring (but not compressing it) and with the 5 kg block in contact with the velocity of the bullet 7 kg block. The 7 kg block is then moved to Answer in units of m/s. the left, compressing the spring a distance 008 (part 1 of 2) 10.0 points 0.7 m, and held in place while the 5 kg block A 1.8 kg steel ball and 3 m cord of negligee remains at rest as shown below. bled mass make up a simple pendulum that can pivot without friction about the point o 0.7 m This pendulum is released from rest in a b zonal position and when the ball is at its lowest point it strikes a 1.8 kg block sitting.

Explanation / Answer

momentum is conserved before and after collission

we take the initial direction of 7 kg mass as the +x direction

momentum along x before collison

       7 Kg             Px   = 7*9.2 = 64.4 kg-m/s

                           Py =0

      19 kg mass   Px = 19*11Cos(22) = 193.78 kg-m/s

                           Py   = 19*11Sin(22)   =   78.29 kg-m/s

momentum after collision

                          Px = 193.78 + 64.4 = 258.18 kg-m/s

                          Py = 78.29 kg-m/s

magnitude = sqrt(258.182 + 78.292) = 269.79 kg-m/s

total mass = 19 +7 = 26 kg

speed after collision = 269.79 /26 = 10.38 m/s

direction tan(p) = 78.29/258.18 = 0.303

p = 16.870 wrt to the +x, axis

The combined mass moves at 16.870 wrt to the +x-axis, which is the original direction od the lighter mass 7 kg

The chnage is direction experienced by the lighter mass is 16.87o

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