Question

The drawing shows two frictionless inclines that begin at ground level (h = 0 m) and slope upward at the same angle θ. One track is longer than the other, however. Identical blocks are projected up each track with the same initial speed v0. On the longer track the block slides upward until it reaches a maximum height H above the ground. On the shorter track the block slides upward, flies off the end of the track at a height H1 above the ground, and then follows the familiar parabolic trajectory of projectile motion. At the highest point of this trajectory, the block is a height H2 above the end of the track. The initial total mechanical energy of each block is the same and is all kinetic energy. The initial speed of each block is v0 = 7.32 m/s, and each incline slopes upward at an angle of θ = 50.0°. The block on the shorter track leaves the track at a height of H1 = 1.25 m above the ground. Find (a) the height H for the block on the longer track and (b) the total height H1 + H2 for the block on the shorter track.

Answer 1

(a) using conservation of energy for longer track

mgH = 1/2mv_o²

H = v_o² / 2g

H = 7.32² / 19.6

H = 2.733 m

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(b)

for shorter track, we find the velocity at the top

v = sqrt ( v_o² - 2gH₁)

v = sqrt ( 7.32² - 2 * 9.8 * 1.25)

v = 5.3928 m/s

Now, use conservation of energy

mg (H1 + H2) + 1/2mv_t² = mgH1 + 1/2mv²

where v_t = v cos 50

so,

solve for H1 + H2

we get,

H1 + H2 = v² + 2gH1 - v_t² / 2g

H1 + H2 = 5.3928² + 2 * 9.8 * 1.25 - 3.4664² / 2 * 9.8

H1 + H2 = 2.12 m