session.masteringphysics.com CH0 Problem 7.42 Resources « previous | 16 of 21 ne
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session.masteringphysics.com CH0 Problem 7.42 Resources « previous | 16 of 21 next » Problem 7.42 Part A A slingshot obeying Hooke's law is used to launch pebbles vertically into the air. You observe that if you pull a pebble back 10.4 cm against the elastic band, the pebble goes 5.9 m high. Assuming that air drag is negligible, how high will the pebble go if you pull it back 20.8 cm instead? Express your answer to two significant figures and include the appropriate units. |h=1 Value Units Submit My Answers Give Up Incorrect; Try Again; 4 attempts remaining Part B How far must you pull it back so it will reach 11.8 m? Express your answer to three significant figures and include the appropriate units s 14.7 cm Submit My Answers Give Up Correct Part C If you pull a pebble that is twice as heavy back 10.4 cm, how high will it go? Fynress voiur answer tn twn sinnificant fiaures and inelule the annronriate unitsExplanation / Answer
Since it is launched vertically, so by conservation of energy, 1/2 k s2 = Mgh
where M= mass of pebble, g = gravity, and k = spring constant, s = pulling back distance which is equivalent to stretching a spring, h = height up to which pebble goes.
10.4cm = 10.4 X .01 m
Therefore, by given values, 1/2 k (10.4 X .01)2 = Mg (5.9) - (a)
(A) Now, we need to find H for x = (20.8 X 0.01)
hence, 1/2 k (20.8 X 0.01)2 = Mg (H) - (b)
deviding eqn. (a) by (b), we get (10.4 X .01)2 / (20.8 X 0.01)2 = 5.9/H
Hence, H = 5.9 X (20.8 X 0.01)2/(10.4 X 0.01)2 = 5.9 X 20.8 X 20.8/ (10.4 X10.4) = 23.6m is answer.
(B) Now, we need to find s for H = (11.8)
hence, 1/2 k (x)2 = Mg (11.8) - (b)
deviding eqn. (a) by (b), we get (10.4 x .01)2 / ( s X 0.01)2 = 5.9/11.8
Hence, s = 14.7 cm is answer.
(c) from the first relation you can see, mass of pebble X height of pebble(H) is directly proportional to s2
hence if mass is doubled, s is kept same then H will get halved as MgH = constant (since s is same as previous, g is always constant)
hence in this case if M doubles, H gets halved. i.e. H = 5.9/2 m = 2.95m
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