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Unformatted text preview: SE 101B/MAE 130B Dynamics Mock Mid-term Examination WI08 SOLUTIONS PROBLEM 1. A 3 lbf ball is given an initial velocity vA of 8 ft/s at an angle of 45O from the horizontal at position A, where the two attached springs are unstretched with free lengths of 12 in. The ball follows the dashed path in space and at some point in the future crosses point B, which is 5 in directly beneath point A. Calculate the velocity vB of the ball at B. Each spring has a stiffness of 10 lbf/in. vA A 12 in 12 in 5 in B SOLUTION: This problem should be solved by energy methods, since all forces acting are conservative. Thus, energy is conserved. This also means that the path that the ball takes through space doesn't matter: the only important information is contained at the endpoints! If we take the lower point (B) to be a gravitational potential energy reference, the ball only has potential and kinetic energy at A (springs are unstretched at A), and this is translated into elastic potential and kinetic energy at B. By using right triangle geometry, the springs stretch to 13 in at B (122 + 52 = 132). So, we have T1 + Vg,1 + Ve,1 = T2 + Vg,2 + Ve,2 1 2 1 1 2 1 2 1 2 1 2 2 mv A + mgh A + k (!x A ) + k (!x A ) = mv B + mghB + k (!x B ) + k (!x B ) 2 2 2 2 2 2 " 5 in % 1 " 3 lbf % 2 1 " 3 lbf % 2 " 3 lbf % 2 " 1 ft % 8 ft/s) + $ (32.2 ft/s2 )$ v + 10 lbf/in)(13 in -12 in) $ '= $ ' $ 2 '( 2' 2' B ( # 12 in/ft & 2 # 32.2 ft/s & #12 in & 2 # 32.2 ft/s & # 32.2 ft/s & v B = 8.54 ft/s PROBLEM 2. An unpowered 100 kg missile is launched upward at with an initial velocity vector of 98.1i+98.1j m/s g from point O. At the top of its trajectory, it explodes into two fragments such that a 75 kg fragment receives an impulse in the m2=25 kg i direction and the 25 kg fragment v0 = 98.1i + 98.1j m1=75 kg x, i O receives an equal and opposite impulse in 400 m the +i direction. Assume no impulses occurred in the j direction. When the fragments hit the ground, they are separated by a horizontal (x-direction) distance of 400 m. Find (a) the velocities of each fragment after the explosion and (b) the magnitude of the impulse delivered. Neglect any air resistance. HINT: The trajectory of the mass center throughout the entire flight duration is that of standard parabolic motion. y, j SOLUTION: The missile is undergoing standard projectile motion until it explodes at the top of its parabolic arc. We can write the impulse-momentum equation for each fragment immediately before and after the explosion; since the impulses were only delivered in the x-direction, we'll write this equation for the x-direction only: m1v0,x ! impulse = m1v1,x m2 v0,x + impulse = m2 v2,x Note these equations may be summed to get a direct expression of conservation of momentum for the entire system (since the impulses were equal and opposite): ( m1 + m2 )v0,x = m1v1,x + m2 v2,x So far, we have two independent equations (the third one just written, remember, is just the sum of the first two, so it's not independent) and 3 unknowns: v1,x, v2,x, and the impulse. We can get further information from the hint: the center of mass of the system continues to follow a parabolic path (as if an explosion never occurred)! We know the equations of motion for this situation, for each direction, are x(t) = v0,x t = 98.1t 1 y(t) = v0,yt ! gt 2 2 1 = 98.1t ! gt 2 2 We can use the second equation here to find the time of flight; at the end of its arc, y = 0, and from that we can find that tflight = 20 s (the other solution is t = 0, but that obviously corresponds to the beginning of the arc, not the end), where the explosion (and thus the separation of the fragments) occurred at 10 s. Now, we can use the bit of information about the final separation of the fragments. We know that x2 x1 = 400 at tflight = 20 s, where x measures the horizontal distance from the origin. Each fragment traveled the first half of the flight at horizontal speed v0,x = 98.1 m/s (when they were stuck together as a single projectile) and the second half of the flight at their respective new horizontal speeds v1,x and v2,x (which we don't know yet). Using the horizontal equation of motion above, we can write ! t flight $ ! t flight $ x1 = v0,x # & + v1,x # & " 2 % " 2 % = 98.1(10) + v1,x (10) ! t flight $ ! t flight $ x2 = v0,x # & + v2,x # & " 2 % " 2 % = 98.1(10) + v2,x (10) but since x2 x1 = 400, we can combine these to get " t flight % " t flight % x2 ! x1 = v2,x $ ' ! v1,x $ ' # 2 & # 2 & 400 = 10v2,x !10v1,x This gives us the necessary third, independent equation, so we can directly plug in to find (a) that v1,x = 88.1 m/s, v2,x = 128.1 m/s, and (b) impulse = 750 kg-m/s. ...
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This note was uploaded on 05/14/2008 for the course SE 101B taught by Professor Todd during the Winter '08 term at UCSD.
- Winter '08