You are given a spring-loaded pellet launcher to play with. The way it works is,

Question

You are given a spring-loaded pellet launcher to play with. The way it works is, you load a pellet of mass .15 kg into the launcher, where it rests against a spring, whose spring constant is k = 2000 N/m.   You pull on a lever which compresses the spring (the amount of compression depends on how hard you pull on the lever). Then you release the lever, which releases the spring back to its relaxed length, and pushes the pellet out. In all that follows, assume gravity is g = 10 m/s2 , and ignore the effects of air resistance.

(a) Suppose you aim the pellet launcher straight up (vertically), and pull the lever back to compress the spring by 10 cm with respect to its relaxed length. You then release the lever, launching the pellet. What maximum height (compared to its release point) will the pellet reach before falling back down again?

(b) Now, suppose that you aim the pellet launcher horizontally, starting from a height of 2.0 m above the ground. You pull the lever back and compress the spring by the same amount as in part (a) above, namely 10 cm. You then release the lever, launching the pellet. How far away will the pellet land from where it was launched (considering only the horizontal displacement)?

(c) Let

Explanation / Answer

a) Ei = Ef

1/2 k x^2 = m g h

0.5*2000*.1^2 = .15*10*h

h=6.67 m

b) first find speed it has after being released

1/2 kx^2 = 1/2 mv^2

2000*.1^2 = .15*v^2

v=11.5 m/s

so y direction, y = y0 + v0y t + 1/2 a t^2

0 = 2 - 0.5*10*t^2

t=0.63 s

so x = v t = 11.5*0.63= 7.25 m

c) now y direction

y = y0 + vy t + 1/2 a t^2

0 = 2 + 11.5*sin(40 degrees)*t - 0.5*10*t^2

t=1.71 s

x direction: x = vx t = 11.5*cos(40 degrees)*1.71= 15.06 m

d) now 1/2 kx^2 = m gh

so doubling m, will halve h, so h =3.34 m

e) doubling m will 1/sqrt(2) v

so distance will be 1/sqrt(2)

so 7.25/sqrt(2)= 5.13 m

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