For this week\'s physics lab, the experimental setup consists of two gliders on

Question

For this week's physics lab, the experimental setup consists of two gliders on a horizontal frictionless air track (see figure below). Each glider supports a strong magnet centered on top of it, and the magnets are oriented so they attract each other. The mass of glider 1 and its magnet is 0.099 kg, and the mass of glider 2 and its magnet is 0.186 kg. You and your lab partners are instructed to take the origin to be at the left end of the track and to center glider 1 at x1 = 0.104 m and glider 2 at x2 = 1.440 m. Glider 1 is 11.1 cm long, and glider 2 is 20.8 cm long, and each glider has its center of mass at its geometric center. When the two gliders are released from rest, they will move toward each other and stick. (a) Predict the position of the center of each glider when they first touch. (Enter your answers from smallest to largest.) x1 = __ m x2 = __ m (b) Predict the velocity that the two gliders will continue to move with after they stick. __ m/s Explain the reasoning behind this prediction for your lab partners. ?

Explanation / Answer

Initial stage: the centre of mass of the system of the two gliders is given as      Xcm = m1 x1 + m2 x2 / (m1 + m2 )             = (0.099)(0.104) + (0.186)(1.44) / (0.099 + 0.186)             = 0.9759m Final stage: When they skck together their center of masses of the gliders are separted by adistance of           X2 -X1 = (1/2)(11.1cm+20.8cm) = 15.95cm or 0.1595m ....... (1) Centre of mass relative to the coordinates of the center of the two gliders is        0.9759 = (0.099)X1 + (0.186)X2 / (0.099+0.186)          0.099X1 + 0.186 X2 = 0.27813 ........ (2)    Solving equation (1) & (2) for X1 and X2      X1 =0.8718m &    x2 = 1.0313m ----------------------------------------------------------------------------------------- NO, the two gliders will not move after they stick. the momentum is conserved. , so it must be zero. Initial stage: the centre of mass of the system of the two gliders is given as      Xcm = m1 x1 + m2 x2 / (m1 + m2 )             = (0.099)(0.104) + (0.186)(1.44) / (0.099 + 0.186)             = 0.9759m Final stage: When they skck together their center of masses of the gliders are separted by adistance of           X2 -X1 = (1/2)(11.1cm+20.8cm) = 15.95cm or 0.1595m ....... (1) Centre of mass relative to the coordinates of the center of the two gliders is        0.9759 = (0.099)X1 + (0.186)X2 / (0.099+0.186)          0.099X1 + 0.186 X2 = 0.27813 ........ (2)    Solving equation (1) & (2) for X1 and X2      X1 =0.8718m &    x2 = 1.0313m ----------------------------------------------------------------------------------------- NO, the two gliders will not move after they stick. the momentum is conserved. , so it must be zero.

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