A cannon elevated at some angle ? (unknown) fires a shell with some initial spee
ID: 2142963 • Letter: A
Question
A cannon elevated at some angle ? (unknown) fires a shell with some initial speed v0 (unknown) that has a range R when fired over level ground (see Figure (a)). During 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, 14% of the shell (m2) continues onward and the other fragment (m1) falls straight down. Determine the distanceD (in terms of R) that the forward moving fragment lands from the cannon (see Figure (b)).
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A cannon elevated at some angle ? (unknown) fires a shell with some initial speed v0 (unknown) that has a range R when fired over level ground (see Figure (a)). During 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, 14% of the shell (m2) continues onward and the other fragment (m1) falls straight down. Determine the distanceD (in terms of R) that the forward moving fragment lands from the cannon (see Figure (b)).Explanation / Answer
let M is the mass of the cannon and theta is the alnge of projection
vo is the initial velocity.
at top point vx = V*cos(theta)
Range, R = vo^2*sin(2*theta)/g
in x - direction
momentum before splitting = momentum ofter splitting
M*vo*cos(theta) = m1*v1 + m2*v2
M*vo*cos(theta) = (86/100)*M*0 + (14/100)*M*v2
v2 = (100/14)*vo*cos(theta)
v2 = 7.143*vo*cos(theta)
time taken to fall down, t = vo*sin(theta)/g
x = v2*t = 7.143*vo*cos(theta)*vo*sin(theta)/g
x = 3.572*vo^2*sin(2*theta)/g
x = 3.572*R
D = R/2 + x = R/2 + 3.572*R = 4.075*R
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