Part D Exercise 8.48 Calculate the change in momentum (that is,the momentum afte
ID: 2281624 • Letter: P
Question
Part D Exercise 8.48 Calculate the change in momentum (that is,the momentum after the collision minus A 100 g marble slides to the left with a velocity of magnitude 0.400 m/s on the frictionless, horizontal surface of an icy New York sidewalk and has a head-on. elastic the momentum before the collision) for 300-g marble. collision with a larger 30.0 g marble sliding to the right with a velocity of magnitude 0200 m/s. Let tzbe to the right. (S the collision is head-on,all the motion is api" ince along a line.) Submit Give Up 0.200 m/s Part E 0400 mls 300g Calculate the change in momentum for 10.0 g marble. Submit Give Up a Tap image to zoom Part F Part A Find the magnitude of the velocity of 30.0-g marble after the collision. This question will be shown after you complete previous question(s. Give Up Submit Part G Calculate the change in kinetic enerzy (that is, the kinetic enery after the collision minus the kinetic energy before the collision) for 300g marble. Part B Find the magnitude of the velocity of IO/0 g marble after the collision. AK Give Up Submit Submit Give U Part H Calculate the change in kinetic energ for 10.0-g marble. Part C Find the direction of the velocity of each marble after the collision. AK Give Up The 100 g marble is moving to the left and the 30.0 g Submit marble is moving to the right. Part I The 300g marble is moving to the left and the 10.0 g g marble is moving to the right. This question will be shown after you complete previous question(s. (O Both marbles are moving to the right. Continue Both marbles are moving to the left.Explanation / Answer
a)from the conservation of momentum
Mv - mv1 = (M+m)V1
30*0.200 - 10*0.400 = 40V1
6 - 4 = 40V1
V1 = 0.05m/s to the right
B)same as A
(C) both marbles are moving to the right
D) Mv - MV1 = M(0.200 - 0.05) = 4.5m/s
E) mv+mV1 = m(v+v1) = 4.5m/s
(g) change in ke = 1/2 * 0.03(0.05^2 - 0.2^2) = -5.625*10^-4 J
(H) CHANGE IN KE = 1/2*0.01(0.05^2 - 0.4^2) = -7.875*10^-4 J
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