(1). An object at rest on a flat, horizontal surface explodes into two fragments

Question

(1). 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 7.10 m before stopping. How far does the lighter fragment slide?Assume that both fragments have the same coefficient of kinetic friction.

(2). A spaceship of mass 1.90×106 kg is cruising at a speed of 6.00×106 m/s when the antimatter reactor fails, blowing the ship into three pieces. One section, having a mass of 4.90×105 kg , is blown straight backward with a speed of 2.50×106 m/s . A second piece, with mass 8.50×105 kg , continues forward at 1.10×106 m/s .What is the speed of the third piece?

Explanation / Answer

(1) Here, apply conservation of momentum -

0 = m * v1 + 7m * v2 ;

=> v1 = -7v2 ;

=> v1 / v2 = -7 ;

Again -

1/2 m v1^2 = mg x ;

so v1^2 = 2 g x1 ; ------1

=> v2^2 = 2 g x2 ; --------2

1 by 2 gives ,

( v1 / v2 )^2 = x1 / x2 ;

7^2 = x1 / x2 = x1 / 7.1

=> x1 = 7.1 x 49 = 347.9 m

So, the distance slide by the lighter fragment = 347.9 m.

(2) First determine the mass of the third piece -

m3=M-(m1+m2) = 1.90x10^6 – (4.9x10^5 + 8.5x10^5) = 5.60 x 10^5 kg.

Now, consider straight Forward direction assumed as positive,

Apply conservation of momentum

mv=Mvo

(1.9x106 kg)(6.0x106 m/s) = -(4.9x105 kg)(2.5x106 m/s)+(8.5x105 kg)(1.1x106 m/s)+(5.6x105 kg)(v m/s)

=> 11.4x10^12 = -12.25x10^11 + 9.35x10^11 + (5.6x105 kg)(v m/s)

=> 11.4x10^12 = -2.9x10^11 + (5.6x105 kg)(v m/s)

=> (5.6x105 kg)(v m/s) = 11.69x10^12

=> v = 2.09x10^7 m/s.

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