A toy gun uses a spring to shoot a ball with a mass of 5.0 g. The spring is “loaded” by compressing it 5.0 cm and
has a spring constant of 288 N/m. The end of the spring is at the end of the gun (the ball does not travel any
distance inside the gun after the spring) and the toy gun is held at a height of 1.5 m above the ground. If the gun is aimed horizontally, how far can it shoot the ball? If it is aimed up 45 degrees how will the initial velocity change? How will the initial velocity change if shot straight upward, and how high would it go?
Why did the toy gun go to spring school?
Because it wanted to get a degree in ballistics!
Now, let's spring into action and calculate the answers to your questions.
First, let's find the potential energy stored in the spring when it's compressed. The formula for potential energy stored in a spring is given by:
PE = (1/2)kx^2,
where PE is the potential energy, k is the spring constant, and x is the displacement (in meters) the spring is compressed.
Plugging in the values:
k = 288 N/m and x = 0.05 m (converted from 5.0 cm),
PE = (1/2)(288 N/m)(0.05 m)^2 = 0.72 J.
Now, let's determine the initial velocity of the ball when shot horizontally. The potential energy of the spring will be converted into the kinetic energy of the ball.
The formula for kinetic energy is given by:
KE = (1/2)mv^2,
where KE is the kinetic energy, m is the mass of the ball (0.005 kg), and v is the initial velocity.
Equating the potential energy to the kinetic energy:
0.72 J = (1/2)(0.005 kg)v^2.
Solving for v:
v^2 = (2*0.72 J) / 0.005 kg
v^2 = 288 m^2/s^2
v = sqrt(288) m/s
v ≈ 16.97 m/s.
So, the initial velocity of the ball when shot horizontally is approximately 16.97 m/s.
Now, let's calculate how far the ball can travel horizontally. Since the ball is shot horizontally, the vertical component of its velocity is zero. The horizontal distance traveled by the ball can be found using the equation:
d = v * t,
where d is the distance, v is the initial horizontal velocity, and t is the time of flight.
Since there is no vertical motion, the time of flight can be determined using the equation:
h = (1/2)gt^2,
where h is the vertical displacement (1.5 m) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Rearranging the equation to solve for t:
t = sqrt((2h) / g)
t = sqrt((2*1.5 m) / 9.8 m/s^2)
t ≈ 0.55 s.
Therefore, the horizontal distance the ball can travel is:
d = (16.97 m/s) * (0.55 s)
d ≈ 9.33 m.
The ball can travel approximately 9.33 meters horizontally when shot from the toy gun.
Now, let's move on to aiming the gun up at a 45-degree angle. When the gun is aimed at an angle, the initial velocity will have both horizontal and vertical components.
The horizontal component of the initial velocity remains the same as when shot horizontally, which is approximately 16.97 m/s.
The vertical component, however, will change by the following amount:
Vertical component = sin(45 degrees) * initial velocity
= sin(45 degrees) * (16.97 m/s)
≈ 12.00 m/s.
So, when aimed up at a 45-degree angle, the initial velocity of the ball will have a vertical component of approximately 12.00 m/s.
If the ball is shot straight upward, the initial velocity will only have a vertical component. The magnitude of the vertical component will remain the same as when shot at a 45-degree angle, which is approximately 12.00 m/s.
Finally, let's calculate the maximum height the ball will reach when shot straight upward.
Using the formula:
v_final^2 = v_initial^2 + 2gh,
where v_initial is the initial velocity, v_final is the final velocity (which becomes 0 at the highest point), g is the acceleration due to gravity, and h is the maximum height.
Rearranging the equation to solve for h:
h = (v_final^2 - v_initial^2) / (2g)
Substituting the values:
h = (0 - (12.00 m/s)^2) / (2*(-9.8 m/s^2))
h ≈ 7.35 m.
Therefore, the ball will reach a maximum height of approximately 7.35 meters when shot straight upward.
That's all for our spring-loaded adventure! Remember to aim high and always clown around responsibly!
To find out how far the toy gun can shoot the ball, we can use the concept of mechanical energy conservation. At the moment the spring is released, all of its potential energy is converted into kinetic energy of the ball. We can set up the equation as follows:
Elastic Potential Energy = Kinetic Energy
1/2kx^2 = 1/2mv^2
Where k is the spring constant, x is the compression distance, m is the mass of the ball, and v is the velocity of the ball.
1/2 * 288 N/m * (0.05 m)^2 = 1/2 * 0.005 kg * v^2
To solve for v, we rearrange the equation:
v^2 = (288 N/m * (0.05 m)^2) / (0.005 kg)
v^2 = 288 N/m * 0.0025 m^2 / 0.005 kg
v^2 = 0.288 m^2/s^2
v ≈ 0.537 m/s
Since the gun is aimed horizontally, the horizontal velocity remains constant throughout the motion. Therefore, the distance it can shoot the ball horizontally can be calculated using the formula:
Distance = Horizontal Velocity * Time
The time can be found using the equation:
Vertical Distance = 1/2 * g * t^2
Where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t is the time of flight.
Since the toy gun is held at a height of 1.5 m above the ground, the vertical distance traveled by the ball can be calculated as:
1.5 m = 1/2 * 9.8 m/s^2 * t^2
Rearranging the equation:
t^2 = (2 * 1.5 m) / 9.8 m/s^2
t^2 ≈ 0.306 s^2
t ≈ 0.553 s
Now we can calculate the horizontal distance:
Distance = 0.537 m/s * 0.553 s
Distance ≈ 0.297 m
Therefore, the toy gun can shoot the ball horizontally for a distance of approximately 0.297 meters.
If the gun is aimed at a 45-degree angle, the initial velocity will change. The initial velocity can be broken down into its horizontal and vertical components using trigonometry. The horizontal component remains the same as before (0.537 m/s), while the vertical component can be found using the formula:
Vertical Velocity = Initial Velocity * sin(θ)
where θ is the angle of 45 degrees.
Vertical Velocity = 0.537 m/s * sin(45°)
Vertical Velocity ≈ 0.382 m/s
Therefore, the initial velocity when aimed at a 45-degree angle would have a magnitude of approximately 0.537 m/s and a vertical component of approximately 0.382 m/s.
If shot straight upward, the horizontal velocity component would be zero, and only the vertical component would be present. The initial velocity would be the same as before (0.537 m/s) but with a positive direction. As for how high the ball would go, we can use the equation for projectile motion:
Vertical Distance = Initial Vertical Velocity * Time - 1/2 * g * t^2
At the highest point, the vertical velocity becomes zero. Thus, we have:
0 = 0.382 m/s - 9.8 m/s^2 * t^2
Solving for t:
t^2 = 0.382 m/s / (9.8 m/s^2)
t^2 ≈ 0.039 s^2
t ≈ 0.197 s
Now we can find the maximum height reached using:
Vertical Distance = Initial Vertical Velocity * Time - 1/2 * g * t^2
Vertical Distance = 0.382 m/s * 0.197 s - 1/2 * 9.8 m/s^2 * (0.197 s)^2
Vertical Distance ≈ 0.075 m
Therefore, if shot straight upward, the ball would reach approximately 0.075 meters in height.
To find the distance the toy gun can shoot the ball when aimed horizontally, we can use the principle of conservation of energy.
First, let's find the potential energy stored in the compressed spring. The potential energy stored in a spring is given by:
PE = (1/2)kx^2
where k is the spring constant and x is the compression of the spring. Plugging in the values, we have:
PE = (1/2)(288 N/m)(0.05 m)^2
= 0.36 J
Since the potential energy of the spring will be converted into kinetic energy of the ball, we can equate them:
KE = PE
The kinetic energy is given by:
KE = (1/2)mv^2
where m is the mass of the ball and v is its velocity. Plugging in the values, we have:
(1/2)(0.005 kg)v^2 = 0.36 J
Simplifying the equation, we get:
v^2 = (2 * 0.36 J) / 0.005 kg
v^2 = 72 m^2/s^2
Taking the square root of both sides, we find:
v = √(72 m^2/s^2)
v ≈ 8.49 m/s
Now, to find the distance the ball can travel, we can use the formula for horizontal projectile motion:
d = v * t
Since there is no initial vertical velocity and the ball is aimed horizontally, the time of flight (t) will be the same as the time it takes for the ball to fall from a height of 1.5 m. Using the equation for free fall, we have:
h = (1/2)gt^2
Plugging in the values, we get:
1.5 m = (1/2)(9.8 m/s^2)t^2
Solving for t, we find:
t^2 = (2 * 1.5 m) / 9.8 m/s^2
t^2 ≈ 0.30612 s^2
Taking the square root of both sides, we find:
t ≈ 0.553 s
Finally, substituting the calculated values, we have:
d ≈ (8.49 m/s) * (0.553 s)
d ≈ 4.70 m
Therefore, when aimed horizontally, the toy gun can shoot the ball approximately 4.70 meters.
Now let's consider the case when the gun is aimed up 45 degrees. In this case, the initial velocity will have both horizontal and vertical components. The horizontal component will remain the same as before (8.49 m/s), but the vertical component will change.
The initial vertical velocity (Vy) is given by:
Vy = V * sin(45°)
where V is the magnitude of the initial velocity (which remains the same).
Therefore, Vy = 8.49 m/s * sin(45°) ≈ 6.00 m/s.
To find the maximum height the ball will reach, we can use the equation for vertical projectile motion:
h = (Vy^2) / (2g)
where g is the acceleration due to gravity (9.8 m/s^2). Plugging in the values, we have:
h = (6.00 m/s)^2 / (2 * 9.8 m/s^2)
h ≈ 1.84 m
So, when shot at an angle of 45 degrees, the ball will reach a maximum height of approximately 1.84 meters.
If the ball is shot straight upward, the initial velocity will be the same as before, but now it will only have a vertical component. Therefore, the initial velocity will be:
V0 = V * cos(45°)
Substituting the value, we have:
V0 = 8.49 m/s * cos(45°) ≈ 6.00 m/s
The maximum height the ball will reach can be found using the equation we used earlier:
h = (V0^2) / (2g)
Plugging in the values, we have:
h = (6.00 m/s)^2 / (2 * 9.8 m/s^2)
h ≈ 1.84 m
Therefore, when shot straight upward, the ball will reach a maximum height of approximately 1.84 meters.