A grenade that is falling vertically explodes into two equal fragments when it is at a height of 2000 m and has a downward velocity of 60 m/s. Immediately after the explosion one of the fragments is moving downward at 80 m/s. Find the position of the second fragment and the position of the center of mass of the system 10 s after the explosion.
There is no new force on the center of mass of the system so it keeps accelerating downward at 9.8 m/s^2 so do the second part first.
h = 2000 - 60(10) - 4.9(100)
First part, let m1 = 1 kg and m2 = 1 kg
before
p = (m1+m2)60 = 120 kg m/s
after
p = m1 V1 + m2 V2 = V1 + 80
p before = p after
120 = V1 + 80
V1 = 40 m/s
so
we have m1 at 2000 meters moving at 40 m/s down and accelerating down at 9.8 m/s^2
h = 2000 - 40(10) - (4.9)(100)
h = 2000 - 400 - 490
thank you so much !
To solve this problem, we can use kinematic equations of motion to find the position of the second fragment and the center of mass of the system.
Step 1: Find the velocity of the second fragment immediately after the explosion.
Given:
Initial downwards velocity of the grenade, v1 initial = -60 m/s
Downwards velocity of the second fragment after the explosion, v2 = -80 m/s
Using the conservation of momentum, we can find the velocity of the second fragment before the explosion.
The total initial momentum of the system is zero (because no external forces are acting on the system).
So, (mass of grenade) * v1 initial = (mass of the first fragment) * (velocity of the first fragment) + (mass of the second fragment) * (-80 m/s)
Since the grenade explodes into two equal fragments, the mass of the first fragment = mass of the second fragment.
Therefore, -60 m/s = (mass of one fragment) * (velocity of one fragment) + (mass of one fragment) * (-80 m/s)
-60 m/s = -140 m/s * (mass of one fragment)
Dividing both sides by -140 m/s, we find:
mass of one fragment = 60 m/s / 140 m/s = 0.4286 kg
Step 2: Find the position of the second fragment 10 s after the explosion.
Using the equation for displacement, we have:
Displacement = Initial velocity * time + (1/2) * acceleration * time^2
Given:
Initial downwards velocity of the second fragment, v2 initial = -80 m/s
Time, t = 10 s
Acceleration due to gravity, g = 9.8 m/s^2 (downwards)
Using these values, we can find the position of the second fragment:
Displacement = -80 m/s * 10 s + (1/2) * 9.8 m/s^2 * (10 s)^2
Simplifying this equation, we find:
Displacement = -800 m + (1/2) * 9.8 m/s^2 * 100 s^2
Displacement = -800 m + 490 m
Displacement = -310 m
Therefore, the position of the second fragment 10 s after the explosion is 310 meters below the point of explosion.
Step 3: Find the position of the center of mass of the system 10 s after the explosion.
Since the mass of the two fragments is equal, the center of mass is located at the midpoint between the positions of the two fragments.
To find the position of the center of mass, we calculate the average of the positions of the two fragments:
Position of the center of mass = (Position of the first fragment + Position of the second fragment) / 2
Using the positions we found earlier, we have:
Position of the center of mass = (2000 m + (-310 m)) / 2
Position of the center of mass = 1690 m / 2
Position of the center of mass = 845 m
Therefore, the position of the center of mass of the system 10 s after the explosion is 845 meters below the point of explosion.
To solve this problem, we need to use the principles of conservation of momentum and conservation of energy.
First, let's find the velocities of the two fragments immediately after the explosion. We are given that the first fragment has a downward velocity of 80 m/s. Since the explosion resulted in two equal fragments, the other fragment must have an upward velocity equal to the negative of the first fragment's velocity. Therefore, the second fragment has an upward velocity of -80 m/s.
Next, we need to determine the time it takes for the first fragment to reach a downward velocity of 80 m/s from a height of 2000 m. We can use the equation of motion:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
In this case, the initial velocity u is 60 m/s (given), the final velocity v is 80 m/s (given), and the acceleration a is due to gravity (-9.8 m/s^2). We can rearrange the equation to solve for time t:
t = (v - u) / a.
Substituting the values, we get:
t = (80 - 60) / -9.8 = 2 seconds.
Therefore, it takes 2 seconds for the first fragment to reach a velocity of 80 m/s.
Now, let's find the position (height) of the second fragment at this time. We can use the equation of motion:
s = ut + (1/2)at^2,
where s is the displacement (height), u is the initial velocity, a is the acceleration, and t is the time.
In this case, the initial velocity u is 0 m/s (since it starts from rest), the acceleration a is due to gravity (-9.8 m/s^2), and the time t is 2 seconds. We can plug in the values and solve for the displacement (height) s:
s = 0 + (1/2)(-9.8)(2)^2 = -19.6 m.
Therefore, the second fragment is at a height of -19.6 m (below the starting point) after 2 seconds.
Finally, to find the position of the center of mass of the system 10 seconds after the explosion, we need to determine the heights of the two fragments at that time.
The first fragment is moving with a velocity of 80 m/s downwards for 10 seconds, so its displacement (height) is given by:
s1 = ut + (1/2)at^2 = 80(10) + (1/2)(-9.8)(10)^2 = 400 - 490 = -90 m.
The second fragment is moving with a velocity of -80 m/s upwards (opposite direction) for 10 seconds, so its displacement (height) is given by:
s2 = ut + (1/2)at^2 = -80(10) + (1/2)(-9.8)(10)^2 = -800 - 490 = -1290 m.
Now, we can find the position of the center of mass using the formula:
x_cm = (m1*x1 + m2*x2) / (m1 + m2),
where m1 and m2 are the masses of the fragments, and x1 and x2 are their respective positions.
Since the two fragments are equal in mass, their masses cancel out in the equation:
x_cm = (x1 + x2) / 2 = (-90 - 1290) / 2 = -690 m.
Therefore, the center of mass of the system is at a height of -690 m 10 seconds after the explosion.