Question

4 Situation 1: On Earth, you kick a box with a speed of 2 ms? up a smooth ramp that makes a 60° angle with the horizontal. Find the maximum dis- tance the box will travel up the ramp. Then find how long before the box reaches you at the base of the ramp again. 1 Situation 2: Now imagine you are on planet X whose mass is twice that of Earth's but with a radius that is also twice that of Earth's. Effectively, this will mean that the acceleration due to gravity near the surface of planet X is 9x = 4.9 ms-2 (roughly half the acceleration due to gravity on Earth). You throw a ball vertically upwards with speed 2 ms? near the surface of planet X. How far up will the ball travel? Then find how long the ball is in the air. Finally, explain in writing why your answers to this calculation on planet X are related to your answers from Situation 1. Neglect friction Situation 3: Now imagine that you kick a box with a speed of 2 ms? up a ramp that makes a 60° with the horizontal, on planet X. Find the maximum distance the box will travel up the ramp. Neglect friction. Situation 4: Then, you move to planet Y, where the acceleration due to gravity is gv = 2.4 ms-2 (roughly half the acceleration due to gravity on planet X). On planet Y, you repeat the experiment, kicking a box with a speed of 2 ms 2 up a ramp that makes a 60° angle with the horizontal. Find the maximum distance up the ramp on planet Y and the time on the ramp. Neglect friction.

Answer 1

Situation ! It o is the angle of incline of the samp the distance the box travel Initial kinetic energy of the box, Ex=22 my² final potential energy, Ep = mgh where his the vertical height Applying law of conservation of energy, Epa Ex = mgh= 1 mv² =) hq= v² 22 =) h=0.204m 29 2x9.8 0.204 sin 60 22 up the ramp is la h = l = 0.236 m sino The maximum distance, l = 0.236m A cideration of the box down lite camp, azg sino = 9-88 S16 60 = 8.49 m/s2 using the relation, Sout+ Į at, laf at² = 0.236 = { x 8:49 xt² ) +² -0.0555 =) t= 0.24 $ Time taken by the box to reach the base of the ramp, t = 0.245 total time for travel, T2 2t² 0.48$ Situation 2 On planet X: 8= 4.9 m/s2, initial speed of throw, u = 2 m/s It is the maximum height the ball attain, Then = 0.408 m 29 2x4.9 The ball will go up a height, h = 0.408 m time taken to reach the highest point, to g Total time spent in ein, To at= ax 0.408=. The height attained in planet x is 2 times that attained on earth because the value of g on planet xů half of that on earth situation 3 : = 6.468m The vertical height reached by the box, h= vz Maximum distance travelled by the box up the ramp, h - l = 0.471m agh=u2 = = u2 u 2 = 0.408 Ş 4.9 0.82 $ 22 ag 2x4-9 l= 0.408 sindo sino

2x2.4 situation 4: On planet Y, 92 2.4 m/s² 22 The vertical height reached by the box, hz v2 =0.833 m ag The maximum distance travelled by the box up the ramp, = l = 0.962 m Acceleration down the ramp, a = g sino - 2.4x Sin 60 = 2.08 m/s2 lihat² = t ² = 2h = 2x0.962 = 0.925 =) t=0.962 § total time spent on ramp, Taat: 2x0.962 =) T = 1.925 l= b 0-833 Sin 60 sino 2.08