The third time I am posting this because I am suffering due to this sum. I solved it in a way and different methods have different answers?

Question

I can't seem to understand the correct concept. Will anyone be kind enough to explain it step-wise for me with a solution. I need to ace an upcoming exam. Thank you in advance

Q) A block of mass m = 3 kg starts to slide from rest along a frictionless inclined surface that makes an angle of φ=30 with respect to the horizontal surface. At the lowest points of the inclined surface, a projectile is fired at a speed of v0 = 12 m/s that makes an angle θ=45 with respect to the horizontal. The aim of this problem is to find the time when the projectile will hit the block.

(a) Find the time when the projectile hits the block. (Given, the initial distance between the block and the projectile along the inclined surface at t = 0 is R0 = 10 m)

(b) Find the height h, where the projectile will hit the block.

(c) Now assuming that the projectile hits the block at its maximum range R along the inclined surface, Rmax=?

? · Accepted Answer

I believe your confusion comes from a very poorly constructed problem.

It cannot be consistently solved as stated. 
Here's why
Lets look at the projectile only
Ignoring air resistance, its initial horizontal and vertical velocities are
vx = vy = 12sin45 m/s
so the x position in time of the projectile is
x(t) = t12cos45
the surface of the slope is the y position in time
y(t) = x(t)tan30 = t12cos45(tan30)
the y position of the projectile is
y(t) = t12sin45 - ½(9.81)t²
the projectile will impact the slope when the y positions are identical
t12sin45 - ½(9.81)t² = t12sin45(tan30)
t is common to all terms, so divides out, rearranging gives
12sin45(1 - tan30) = 4.905t
t = 0.731 s
so at impact, the following positions are true
y = 3.58 m = (12cos45)(tan30)(0.731)
x = 6.20 m = (12cos45)(0.731)
s = 7.16 m along ramp from firing position = y/sin30

The block will slide from rest a distance
d = ½(gsinθ)t²
d = ½(9.81sin30)(0.731²) = 1.31 m

so the farthest the block can be from the ramp base is
R = 7.16 + 1.31 = 8.47 m
if fired and released at the same instant.
Telling us that the block starts from 10 m up the ramp guarantees that no solution is possible. Either the block has to be closer if the release time is the same, or a time delay between block release at 10 m and projectile firing is needed.

The time needed for the block to slide from 10 m to 7.12 m is
t = √(2(10 - 7.12)/9.81sin30) = 1.08 s
so the projectile would have to be fired 1.08 - 0.731 = 0.353 s after the block is released.

Please remember to select a "Favorite Answer" from among your results.

Trending News

The third time I am posting this because I am suffering due to this sum. I solved it in a way and different methods have different answers?

1 Answer