A cannon elevated at some angle ? unknown fires a shel with some initial speed v

Question

A cannon elevated at some angle ? unknown fires a shel with some initial speed vo unknown) that has a range R when fired er eve ground (ee Figure ou ng a test fire, a shell explodes at the top of the trajectory into two unequal mass fragments. The shell explodes in such a manner that neither fragment experiences a change in momentum in the y direction. However, 18% of the shell (m2) continues onward and the other fragment (ml) falls straight down. Determine the distance D (in terms of R) that the forward moving fragment lands from the cannon (see Figure (b)) D= Explosion site To (a) No Explosion (b) With Explosion Tutorial

Explanation / Answer

Solve this problem using conservation of momentum -

Horizontal range of object projected at an angle ? is given by –

R = (V²/g)sin(2?)

Where,
V is initial velocity in m/s
g is the acceleration of gravity

Time of flight = R/(Vcos?)

And, maximum height, h = (V²/2g)sin²?

Now, if M1 falls straight down, that means it's horizontal momentum is zero, which means it's momentum due to it's trajectory is canceled. Which means M2 has it's original momentum plus that due to the explosion, which is M1's momentum.

Given that, M1 = (1.0 – 0.18) = 0.82M

And M2 = 0.18M

At the top of the trajectory, speed is equal to initial speed times cos ?

momentum of shell before explosion is MV?cos?
momentum of M1 = 0 after the explosion, which means it loses (0.82)MV?cos? which M2 gains
so momentum of M2 is original momentum (0.18MV?cos?) now has added to it (0.82)MV?cos? for a total of MV?cos?

but the mass is 0.3M so the P = MV?cos? = 0.18MV
where V is the speed after the explosion

So, V = MV?cos? / 0.18M = (50/9)V?cos?

Range is divided in half. First half is as above,
R? = (1/2)(V?²/g)sin(2?)
Second half, because of higher speed, is
R? = (1/2)(((50/9)V?cos?)²/g)sin(2?)
R? = (1/2)(50/9)²cos²?R
R? = (1250/9)cos²?R

Therefore, total distance, D = R/2 + R?

=> D = R((1/2) + (1250/9)cos²?)

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