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Two blocks are free to slide along the frictionless wooden track shown below. Th

ID: 2181205 • Letter: T

Question

Two blocks are free to slide along the frictionless wooden track shown below. The block of mass m1 = 4.93 kg is released from the position shown, at height h = 5.00 m above the flat part of the track. Protruding from its front end is the north pole of a strong magnet, which repels the north pole of an identical magnet embedded in the back end of the block of mass m2 = 10.5 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision.

http://www.google.com/imgres?q=Two+blocks+are+free+to+slide+along+the+frictionless+wooden+track+shown&hl=en&biw=1440&bih=684&tbm=isch&tbnid=MjXnb5HPYMUlUM:&imgrefurl=http://www.chegg.com/homework-help/questions-and-answers/blocks-free-slide-frictionless-wooden-track-shown--block-mass-m1-506-kg-released-position--q1624371&docid=qPEPXhLiuoEOzM&imgurl=https://s3.amazonaws.com/answer-board-image/3257707a-d8e1-4802-8f50-7289f5bdd437.gif&w=420&h=148&ei=fXZ3UJ6wM6f42QXe4YCoBw&zoom=1&iact=rc&dur=290&sig=109620160885161235302&page=1&tbnh=72&tbnw=204&start=0&ndsp=18&ved=1t:429,r:0,s:0,i:70&tx=67&ty=25

Explanation / Answer

u = sqrt(2 * g *h) = sqrt(2 * 9.8 *5) = 9.90 m/s

4.99* 9.90 = 10.7 * v1 - 4.99*v2

49.4 = 10.7 * v1 - 4.99*v2 ---------- (1)

v1 + v2 = 9.90 ------------------- (2)

Solve the system of eqtn, (http://www.webmath.com/solver2.html)

v1 = 6.3 m/s

v2 = 3.6 m/s

mgh = mv2 /2

h = 3.6^2 / (2*9.8) = 0.66 m

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