How to have scrolling vertically as well as horizontally inside fragment of fragment activity?

Momentum friction problem: How far does the lighter fragment slide?

• An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavier fragment slides 6.50 m before stopping. Assume they both have the same coefficient of friction. How far does the lighter fragment slide?

• Answer:

  Using the law of conservation of momentum, MV = M1(V1) + M2(V2) where M = initial mass of the object V = initial velocity of the object = 0 (at rest) M1 = mass of the first fragment V1 = velocity of the first fragment after explosion M2 = mass of the second fragment V2 = velocity of the second fragment after explosion 0 = M1(V1) + M2(V2) since M1 = 7(M2), then the above becomes 7(M2)(V1) + M2(V2) = 0 "M2" will cancel out, hence 7V1 = -V2 ------ call this Equation 1. From this equation, it is noted that the two fragments went into opposite directions after the explosion. The next working equation is Vf^2 - Vi^2 = 2as where Vf = final velocity of the fragment = 0 (since it will stop) Vi = velocity of the fragment after explosion a = acceleration of the fragment s = distance travelled by the fragment before stopping For the first fragment, 0 - (V1)^2 = 2(a)(6.5) -(V1)^2 = 13a -- call this Equation 2 From Newton's 2nd Law of Motion, F = ma where F = frictional force acting on the fragment F = µmg and therefore µmg = ma and solving for "a", a = µg = 9.8µ and substituting this into Equation 2, -(V1)^2 = 13(9.8µ) --- call this Equation 3 For the lighter fragment, 0 - (V2)^2 = 2(a)(S) where a = acceleration of the lighter fragment S = stopping distance of the lighter fragment As derived above, a = µg and the above becomes, -(V2)^2 = (2)(µ)(9.8)S -(V2)^2 = 19.6(µ)S Since V2 = -7V1 (from Equation 1), then the above becomes -(7V1)^2 = 19.6(µ)S --- call this Equation 4 Dividing Equation 4 by Equation 3, -(7V1)^2/-(V1)^2 = 19.6(µ)S/13(9.8µ) 49 = 19.6S/13*9.8 and solving for "S" S = 318.5 m Hope this helps. BTW, please check my arithmetic. It can be out of what at times!!!