Question

I need help....!

A bomb explodes into three identical pieces. One flies in the + y-direction at 17.3 m/s, a second in the - x-direction at 11.6 m/s.

What's the magnitude of the velocity of the third piece?

What's the direction of the velocity of the third piece? Count counterclockwise from the + x axis.

Answer 1

Do you know the Law of conservation of momentum?

Answer 2

yes i kinda do but i don't know where to start momentum=mass*velocity

Answer 3

See, according to law of conservation of momentum, Momentum of a system should be same before and after any event if no external force is applied on that system. Get it?
Final Momentum = Initial Momentum
What is the initial momentum of the system? The bomb was stationary before explosion, and so it is the zero. So, the final momentum should also be zero(a lot of forces act, but not on the bomb, but within it).
List the final momentum of each piece. Take the one unknown to be a variable. Momentum is a vector, so be careful with the directions. Take the positive x-direction to be i-cap, positive-y direction to be j-cap. Now add all these momenta to zero. Find the variable.

Answer 4

Mzy,

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Answer 5

Initially, the bomb is in the rest,due to internel force it explodes ,there is no external force is present here,Hence we apply coservation of liner momentum of the bomb.
Here the bomb is in rest at initilly due to this it has zero linear momentum at initial.Let the mass of the bomb is m.
by the conservation of linear momentum:
initial linear momentum=final liner momentum
0=m*(-11.6i)+m*(17.3)j+m*vector (v)
here i,&j is unit vector along x and y axis respectively.and vector(v) is the velocity of third part.after solve this equation we get:vector(v)=11.6i-17.3j. i.e magnitude of vector v is 51.25m/s and its direction i.e angle=tan inverse of(17.3/11.6) i.e angle in fourth quadrant with x axis in counter clockwise direction is .9035rad.

Answer 6

can u show me how you got 51.25 m/s? i am still a bit confused