On a highly polished, essentially frictionless lunch counter, a 0.410 submarine
ID: 2184267 • Letter: O
Question
On a highly polished, essentially frictionless lunch counter, a 0.410 submarine sandwich moving 2.90 to the left collides with a 0.260 grilled cheese sandwich moving 1.05 to the right.a.)If the two sandwiches stick together, what is their final velocity?
b.)How much mechanical energy dissipates in the collision?
Explanation / Answer
On a greasy, essentially, frictionless lunch counter, a 0.520 kg submarine sandwich, moving 2.40 m/s to the left, collides with an 0.260 kg grilled cheese sandwich moving 1.00 m/s to the right. (a) If the two sandwiches stick together, what is the final velocity, taking positive velocities to the right? m/s (b) How much mechanical energy is dissipated in the collision? J Answers --> (a) -1.27 m/s (so, to the left) (b) Loss of 1.0 J How and why... Inelastic collision (kinetic energy is not conserved) Define right as positive (+). In momentum problems, direction is very important. m1 = 0.520 kg u1 = -2.40 m/s m2 = 0.260 kg u1 = +1.00 m/s Solve for common post-impact velocity v * [m1 + m2] = m1u1 + m2u2 v * [ (0.520 kg) + (0.260 kg) ] = [ (0.520 kg) * (-2.40 m/s) ] + [ (0.260 kg) * (+1.00 m/s) ] v * [ 0.78 kg ] = [ -1.248 kg-m/s ] + [ 0.26 kg-m/s ] v * [ 0.78 kg ] = [ -0.988 kg-m/s ] v = [ -0.988 kg-m/s ] / [ 0.78 kg ] v = -1.27 m/s I assume you mean kinetic energy, not mechanical energy KE = 0.5 * m * v^2 KE1 = 0.5 * (0.52 kg) * (2.40 m/s)^2 KE1 = (0.26 kg) * (5.76 m^2/s^2) KE1 = 1.50 J KE2 = 0.5 * (0.26 kg) * (1.00 m/s)^2 KE2 = (0.13 kg) * (1.0 m^2/s^2) KE2 = 0.13 J Total pre-impact system kinetic energy KEPr = KE1 + KE2 KEPr = (1.50 J) + (0.13 J) KEPr = 1.63 J Post impact system energy KE = 0.5 * [ (0.52 kg) + (0.26 kg) * (1.27 m/s)^2 KE = 0.5 * [ 0.78 kg ] * (1.62 m^2/s^2) KE = 0.63 J So lost KEL = KEPr - KE KEL = (1.63 J) - (0.63 J) KEL = 1.0 J
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