A 15 g ball is shot from a spring gun whose spring constant is . The spring is compressed by 5 cm. The greatest possible horizontal range of the ball for this compression is
Answers
Answer:
The energy stored in the spring is given by:
E =1 /2 k x ^2
k is the force constant. x is the extension.
∴
E = 1 /2 × 600 × [0.05] ^2 = 0.75 x J
I will assume that all this energy will appear as the kinetic energy of the ball:
∴ K E = 1 /2 m v ^2
v = √ 2 K E m
= √ 2 × 0.75 0.015
=10 x m/s
I will assume that the ball is being launched from the ground as no height is given. The range is given by:
d = v ^2 sin 2 θ /g
Where θ is the angle of launch. To find the value which will give the maximum range we find the 1st derivative and set it to zero.
Using The Chain Rule:
d
(
d
)
d
(
θ
)
=
v
2
g
×
2
cos
2
θ
=
0
∴
cos
2
θ
=
0
∴
2
θ
=
π
2
θ
=
π
4
=
45
∘
This is the launch angle that gives the maximum range.
Using this value gives:
d
=
10
2
×
sin
90
10
=
100
×
1
10
=
10
x
m
Explanation: