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Unformatted text preview: Solution Derivations for Capa #9 1) A 48 kg shell is fired from a gun with a muzzle velocity 115 m/s at 55 above the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls vertically. What is the horizontal speed of the other fragment? m = Given v = Given = Given v 1 x = ? Right before the explosion, the velocity of the projectile is only the horizontal component of the initial velocity, or v x = v cos . Momentum is always conserved, so the initial momentum must equal the final momentum. To simplify the prob lem (since momentum is a vector quantity), we can consider only the horizontal momentum. Since the particle splits into two equally massed particles, the final momentum is the sum of momenta of the two particles. So, mv x = 1 2 mv 1 x + 1 2 mv 2 x . Since one particle completely stops, its horizontal velocity is zero. The equation becomes mv x = 1 2 mv 1 x . The m s cancel, and were left with v x = 1 2 v 1 x . So, v 1 x = 2 v x = 2 v cos 2) Two identical steel balls, each of mass 2 . 6 kg , are suspended from strings of length 34 cm so that they touch when in their equilibrium position. We pull one of the balls back until its string makes an angle = 54 with the vertical and let it go. It collides elastically with the other ball. How high will the other ball rise above its starting point? m = Given, equal for both balls. L = Given = Given Since the collision is elastic, the second ball will rise to the same height as the first ball. The height from the ball to the mass can be found from a triangle and 1 the height of the ball can be found by subtracting the previous height from the string height. h = L L cos = L (1 cos ) This can be shown to be true by conservation of energy: KE i + PE i = KE f + PE f Since the masses are equal, 0 + m 1 gh 1 = 0 + m 2 gh 2 mgh 1 = mgh 2 h 1 = h 2 3) Suppose that instead of steel balls we use putty balls . They will collide inelastically and remain stuck together after the collision. How high will the balls rise after the collision? m = Given L = Given = Given Since this collision is inelastic, only conservation of momentum applies. However, we can still use the energy equations but only for the particles before and after (not during) the collision. Momentum is always conserved, so we can always use that equation. In an inelastic collision, m 1 v 1 i + m 2 v 2 i = ( m 1 + m 2 ) v f Here, the masses are the same and the second particle has no initial velocity. mv 1 i = 2 mv f The m s cancel, leaving 1 2 v 1 i = v f (1) In order to get v 1 i , we can use the energy equation 1 2 mv 2 1 i = mgh v 1 i = q 2 gh We found an equation for h in the previous problem, so v 1 i = q 2 gL (1 cos ) 2 Plugging this into (1), we can solve for v f 1 2 v 1 i = v f q 2 gL (1 cos ) 2 = v f Now that we know the final velocity of the two masses when they stick together, we can find the height by using the energy equation once again, this time solving...
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 Spring '08
 SPIKE,BENJ
 Physics, Mass

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