Question

The initial speed of each block is v0 = 6.92 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

Answer:

Part a)

$H = 2.44 m$

Part b)

$H_1 + H_2 = 1.95 m$

Explanation:

Part a)

As we know that there is no friction on the inclined plane

So we can use energy conservation for both planes

For longer plane we will have

$mgH = \frac{1}{2}mv^2$

$H = \frac{v^2}{2g}$

$H = \frac{6.92^2}{2(9.81)}$

$H = 2.44 m$

Part b)

On shorter plane the speed while it leave the plane at height H1 is given as

$\frac{1}{2}m(v_f^2 - v_i^2) = -mgH_1$

$\frac{1}{2}m(v_f^2 - 6.92^2) = -(9.81)(1.25)$

$v_f = 4.83 m/s$

now the maximum height of projectile is given as

$H_2 = \frac{v_f^2 sin^2\theta}{2g}$

$H_2 = \frac{4.83^2sin^2(50)}{2(9.81)}$

$H_2 = 0.70 m$

so we have

$H_1 + H_2 = 1.25 + 0.7$

$H_1 + H_2 = 1.95 m$