A rocket-powered hockey puck has a thrust of 4.70 N and a total mass of 2.40 kg. It is released from rest on a frictionless table, 4.80 m from the edge of a 3.00 m drop. The front of the rocket is pointed directly toward the edge.
How far does the puck land from the base of the table?
To solve this problem, we can first calculate the time it takes for the puck to reach the edge of the table, and then use that time to determine the horizontal distance it travels.
Step 1: Calculate the time to reach the edge of the table
Using the equation of motion for constant acceleration, we can calculate the time taken for the puck to travel 4.80 m horizontally. The equation is:
s = ut + (1/2)at^2
where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.
In this case, the initial velocity is 0 m/s as the puck is released from rest. The acceleration in the horizontal direction (ax) is given by the thrust divided by the mass of the puck:
ax = F / m
ax = 4.70 N / 2.40 kg
Calculate the value of ax:
ax = 1.9583 m/s^2 (rounded to 4 decimal places)
Using this value, we can rearrange the equation of motion to solve for time:
s = (1/2)at^2
4.80 = (1/2)(1.9583)t^2 (Substituting the known values)
Simplify the equation:
9.60 = 0.97915t^2
Divide both sides by 0.97915:
t^2 = 9.8076
Take the square root of both sides to solve for t:
t = √9.8076
t = 3.133 s (rounded to 3 decimal places)
Step 2: Calculate the horizontal distance traveled by the puck
Now that we know the time taken to reach the edge of the table, we can determine the horizontal distance covered by the puck using the equation:
s = ut
where s is the distance, u is the initial velocity, and t is the time.
In this case, the initial velocity (ux) is also 0 m/s as the puck is released from rest. So the equation simplifies to:
s = 0 × 3.133
s = 0 m
Therefore, the puck lands right at the edge of the table, 0 m from the base.
To find how far the rocket-powered hockey puck lands from the base of the table, we can calculate the horizontal distance it will travel before dropping from the table.
First, let's calculate the time it takes for the puck to drop from the table. We can use the formula for the time taken by an object to fall vertically:
t = sqrt(2h/g)
where
t is the time taken
h is the height from which the puck drops
g is the acceleration due to gravity (approximately 9.81 m/s²)
Plugging in the values:
h = 3.00 m
g = 9.81 m/s²
t = sqrt(2 * 3.00 m / 9.81 m/s²)
t ≈ 0.78 s
The puck will take approximately 0.78 seconds to drop from the table.
Next, let's calculate the horizontal distance traveled by the puck during this time. We can use the formula for distance traveled horizontally by an object with constant acceleration:
d = v * t
where
d is the distance traveled
v is the horizontal velocity
t is the time taken
We need to find the horizontal velocity. Since the puck is rocket-powered, we can use Newton's second law of motion to find the acceleration:
F = ma
where
F is the thrust force provided by the rocket
m is the mass of the puck
a is the acceleration
Rearranging the equation, we get:
a = F/m
Plugging in the values:
F = 4.70 N
m = 2.40 kg
a = 4.70 N / 2.40 kg
a ≈ 1.96 m/s²
Now, we can find the horizontal velocity:
v = a * t
Plugging in the values:
a = 1.96 m/s²
t = 0.78 s
v = 1.96 m/s² * 0.78 s
v ≈ 1.53 m/s
Therefore, the horizontal velocity of the puck is approximately 1.53 m/s.
Finally, we can calculate the horizontal distance traveled by the puck:
d = v * t
Plugging in the values:
v = 1.53 m/s
t = 0.78 s
d = 1.53 m/s * 0.78 s
d ≈ 1.19 m
Therefore, the puck will land approximately 1.19 meters from the base of the table.
How long does the rocket fire in time...I assume the entire trip.
F=ma a=4.7/2.4
time to get to the edge:
4.8=1/2 a t^2
t= sqrt 9.6*2.4/4.7 = 2.21sec
time to drop:
t=sqrt(1/2 g *3)=3.83 sec
Velocity at edge= sqrt(2.21*4.7/2.4)=2.08m/s
distancefrom table= vi*t+1/2 at^2
where t is the time to drop
= 2.08*3.83+1/2*4.7/2.4*3.83^2
and you can compute that, check it all.