As viewed from above, a fireworks box with mass m = 6.0 kg slides with speed v =

Question

As viewed from above, a fireworks box with mass m = 6.0 kg slides with speed v = 3.0 m/s across a frictionless floor in the positive x direction. It suddenly explodes into two pieces. One piece, with mass Ma = 2.0 kg, moves off at an angle of theta(a) = 30 degrees above the positive x axis with speed Va= 7.0 m/s.

a. What is the speed of the second piece, with mass Mb?

b.What is the direction of the second piece; i.e. what is the angle theta(b)?

c. What magnitude impluse did the explosion exert on the second piece?

Explanation / Answer

Since there is no external force associated, Conservation of momentum is valid in both x&y direction In x direction, 6*3=(2*7*cos(30))+(4*k) k=1.4683m/s. In y direction 2*7*sin(30)+4*7*l=0 l=-0.25 m/s. which means that the second mass is moving in the fourth quadrant ie.tan^(-1)(0.25/1.4683)=9.66 degrees 1)the speed being 1.4894m/s. 2)the angle is 9.666 degrees in the negative direction to the positive x-axis) 3)m*(dv)=4*(3-1.4894)=6.0424 N-sec

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