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2 years ago

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The spring in the muzzle of a child's spring gun has a spring constant of 730 N/m. To shoot a ball from the gun, first the sprin

g is compressed and then the ball is placed on it. The gun's trigger then releases the spring, which pushes the ball through the muzzle. The ball leaves the spring just as it leaves the outer end of the muzzle. When the gun is inclined upward by 31° to the horizontal, a 53 g ball is shot to a maximum height of 1.90 m above the gun's muzzle. Assume air drag on the ball is negligible. (a) At what speed does the spring launch the ball? (b) Assuming that friction on the ball within the gun can be neglected, find the spring's initial compression distance.

1 answer:

Korolek [52]2 years ago

5 0

Answer:

a. V=11.84 m/s

b.x=0.052m

Explanation:

a).

Given

$K=730 N/m$ , $m=0.053kg$ , $h=1.90m$ .

$v_f^2=v_i^2+2*g*h$

$v_i^2=2*g*h=2*9.8m/s^2*1.9m$

$v_i=\sqrt{2*9.8m/s^2*1.9m}=\sqrt{37.24 m^2/s^2}$

$v_i=6.1 m/s$

$v_i=V*sin(31)$

$V=\frac{v_i}{sin(31)}=\frac{6.1m/s}{sin(31)}$

$V=11.84 m/s$

b).

$K_k=\frac{1}{2}*K*x^2$

No friction on the ball so:

$x^2=\frac{2*K_k}{K}$

$x=\sqrt{\frac{2*0.053kg*9.8m/s^2*1.9m}{730N/m}}$

$x=\sqrt{2.7x10^{-3}m^2}=0.052m$

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