1) A crude cannonball launcher is constructed using a 45? incline and a spring,
ID: 2013994 • Letter: 1
Question
1) A crude cannonball launcher is constructed using a 45? incline and a spring, as shown in Fig.-1. When the spring is compressed, the center of the cannonball is 1.50 m off the ground, as shown in Fig.-1a. When the spring is let go, it expands to its equilibrium position, as shown in Fig.-1b. At this point, the cannonball is just leaving the incline, and its center is 3.00 m off the ground. If the mass of the cannonball is 10.0 kg, and the force constant of the spring is 200 N/m, find the horizontal distance that the cannonball lands from point A. Neglect all friction, and all masses except for the mass of the cannonball.this is the problem that i tried but i dont know how to start it
Explanation / Answer
The cannon ball has an initial velocity of zero then the spring applies a force that can be calculate through Fsp=-kx (the negative value doesn't really apply in this situation)(k is the spring constant, x is the distance streched or compressed)
so the k is given as 200N/m and you must calculate the x. In this situation the give the initial, and final heights of the cannon ball and the angle it is fired. By using the sin function we can say
sin () = 1.5/hyp (the hyp is the x) so
x=1.5/(sin(45)) 2.12m
Now you can calculate
Fsp = 200(2.12) = 424N Now you can find the acceleration of the cannon ball for the compressed distance. Then find the velocity it leaves the spring.
F=ma 424= (10)a a=42.4m/s2
v2=vo2 + 2a(s-so) v2=02 + 2(42.4)(2.12-0) v=13.4 m/s (s is the position)
now that you have the speed it leaves the ramp, we split the motion into the x and y direction. We will use the y direction motion to get the time the cannonball is in the air then use that time with the x direction motion to calculate the distance it travels horizontally.
vx=13.4*cos(45)=9.48m/s vy=13.4*sin(45)=9.48m/s
sy=s0y+v0y(t)+(1/2)(a)(t2) 0=3+9.48(t)+(1/2)(-9.8)(t2) using the quadratic formula (t can only be positive)
t=2.21s
sx=s0x+v0x(t)+(1/2)(a)(t2) sx=0+9.48(2.21)+(1/2)(0)(2.212) sx=9.48(2.21)=20.95m
The cannon ball will land 20.95m away from the point it leaves the ramp. (I'm not sure what point A means)
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