An object with total mass mtotal = 15.8 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.5 kg moves up and to the left at an angle of θ1 = 18° above the –x axis with a speed of v1 = 27.6 m/s. A second piece with mass m2 = 5.4 kg moves down and to the right an angle of θ2 = 23° to the right of the -y axis at a speed of v2 = 21.5 m/s.
a. What is the magnitude of the final momentum of the system (all three pieces)?
b. What is the mass of the third piece?
c. What is the x-component of the velocity of the third piece?
d. What is the y-component of the velocity of the third piece?
e. What is the magnitude of the velocity of the center of mass of the pieces after the collision?
f. Calculate the increase in kinetic energy of the pieces during the explosion.
(a) 0 kg*m/s
(b) (15.8 - 4.5 - 5.4)kg
= 5.9 kg
(c) 0 = [4.5*27.6*cos162º + 5.4*21.5*cos293º + 5.9*V*cosΘ3] kg·m/s
for V in m/s
horizontal component V*cosΘ3 = 72.8/5.9 m/s
= 12.3 m/s
(d) 0 = [4.5*27.6*sin162º + 5.4*21.5*sin293º + 5.9*V*sinΘ3] kg·m/s
for V in m/s
vertical component V*sinΘ3 = 68.4/5.9 m/s
= 11.6 m/s
(e) 0 m/s
(f) 0.5 * [4.5*27.6² + 5.4*21.5² + 5.9*(12.3² + 11.6²)] J
= 3810 J
a. The magnitude of the final momentum of the system (all three pieces) can be found by adding the individual momenta of each piece.
The momentum of the first piece is given by:
p1 = m1 * v1
p1 = 4.5 kg * 27.6 m/s
The momentum of the second piece is given by:
p2 = m2 * v2
p2 = 5.4 kg * 21.5 m/s
The magnitude of the final momentum is then:
p_total = p1 + p2
b. The total mass of the system is given as mtotal = 15.8 kg. Since we know the masses of the first two pieces (m1 = 4.5 kg and m2 = 5.4 kg), we can find the mass of the third piece by subtracting the sum of their masses from the total mass:
m3 = mtotal - (m1 + m2)
c. The x-component of the velocity of the third piece can be found by considering the conservation of momentum in the x-direction. Since the first piece moves up and to the left and the second piece moves down and to the right, the x-component of their momenta will cancel each other out. Therefore, the x-component of the velocity of the third piece will be zero.
d. The y-component of the velocity of the third piece can also be found by considering the conservation of momentum in the y-direction. Since the first piece moves up and the second piece moves down, the y-component of their momenta will again cancel each other out. Therefore, the y-component of the velocity of the third piece will also be zero.
e. The magnitude of the velocity of the center of mass of the pieces after the collision can be found by dividing the total momentum of the system by the total mass of the system. Since the x-component and y-component of the velocity of the third piece are zero, the magnitude of the velocity of the center of mass will also be zero.
f. The increase in kinetic energy of the pieces during the explosion can be calculated by subtracting the initial kinetic energy of the system from the final kinetic energy of the system.
The initial kinetic energy of the system is given by:
KE_initial = 0.5 * mtotal * 0^2
The final kinetic energy of the system is given by:
KE_final = 0.5 * (m1 * v1^2 + m2 * v2^2 + m3 * 0^2)
The increase in kinetic energy is then:
ΔKE = KE_final - KE_initial
To solve this problem, we'll use the conservation of momentum principle. According to this principle, the total momentum before the explosion is equal to the total momentum after the explosion.
a. The magnitude of the final momentum of the system is the sum of the momenta of the individual pieces. We can calculate the momentum of each piece using the formula: momentum = mass * velocity.
Given:
m1 = 4.5 kg
θ1 = 18° (above the –x axis)
v1 = 27.6 m/s
To find the x and y components of the velocity, we'll use the trigonometric functions:
vx1 = v1 * cos(θ1)
vy1 = v1 * sin(θ1)
Substituting the given values, we get:
vx1 = 27.6 * cos(18°)
vy1 = 27.6 * sin(18°)
Similarly, for the second piece, we have:
m2 = 5.4 kg
θ2 = 23° (to the right of the -y axis)
v2 = 21.5 m/s
vx2 = v2 * cos(θ2)
vy2 = v2 * sin(θ2)
Substituting the given values, we get:
vx2 = 21.5 * cos(23°)
vy2 = 21.5 * sin(23°)
Now we can calculate the momentum for each piece:
momentum1 = m1 * sqrt(vx1^2 + vy1^2)
momentum2 = m2 * sqrt(vx2^2 + vy2^2)
Finally, the magnitude of the final momentum is the sum of these two:
final_momentum = momentum1 + momentum2
b. To find the mass of the third piece, we'll subtract the masses of the first two pieces from the total mass:
m3 = mtotal - m1 - m2
c. The x-component of the velocity of the third piece will be the negative of the sum of the x-components of the velocities of the first two pieces:
vx3 = -(vx1 + vx2)
d. The y-component of the velocity of the third piece will be the negative of the sum of the y-components of the velocities of the first two pieces:
vy3 = -(vy1 + vy2)
e. To find the magnitude of the velocity of the center of mass, we can use the formula: velocity_center_of_mass = total_momentum / total_mass.
Substituting the calculated values, we get:
velocity_center_of_mass = final_momentum / (m1 + m2 + m3)
f. The increase in kinetic energy of the pieces during the explosion is given by: increase_in_kinetic_energy = (final_momentum^2 - initial_momentum^2) / (2 * (m1 + m2 + m3))
Now let's substitute the given values and calculate the results.
To solve this problem, we will use the principle of conservation of momentum. The initial momentum of the system is zero since the object is at rest before the explosion. After the explosion, the total momentum of the system will still be conserved.
a. The magnitude of the final momentum of the system can be found by summing up the momentum of each individual piece. The momentum (p) of an object is given by the product of its mass (m) and velocity (v), so we can use the equation p = mv.
For the first piece:
Magnitude of momentum (p1) = m1 * v1 = 4.5 kg * 27.6 m/s
For the second piece:
Magnitude of momentum (p2) = m2 * v2 = 5.4 kg * 21.5 m/s
The third piece has an unknown mass and velocity, so we'll consider it as m3 and v3, respectively.
Since momentum is a vector quantity, we need to consider both magnitude and direction. The momentum of the third piece can be broken down into x and y-components. Let's represent the x-component of the momentum as p3x and the y-component as p3y.
The x-component of momentum is given by:
p3x = m3 * vx
The y-component of momentum is given by:
p3y = m3 * vy
Since the initial momentum of the system is zero, the final x-component and y-component of momentum should also be zero. Therefore, we have the following equations:
p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
We can now proceed to solve the equations to find the unknowns.
b. The mass of the third piece (m3) can be found by rearranging the equation for the x-component of momentum:
m3 = - (p1x + p2x) / vx
c. The x-component of the velocity of the third piece can be found by rearranging the equation for the x-component of momentum:
v3x = p3x / m3
d. The y-component of the velocity of the third piece can be found by rearranging the equation for the y-component of momentum:
v3y = p3y / m3
e. The magnitude of the velocity of the center of mass of the pieces after the collision is given by the equation:
vcm = sqrt[(v1x + v2x + v3x)^2 + (v1y + v2y + v3y)^2]
f. The increase in kinetic energy of the pieces during the explosion can be found by calculating the difference between the initial and final kinetic energies of the system.
Initial kinetic energy of the system = 0 (since the object is at rest)
Final kinetic energy of the system = (1/2) * (m1 * v1^2 + m2 * v2^2 + m3 * v3^2)
Increase in kinetic energy = final kinetic energy - initial kinetic energy
Now, you can substitute the given values and calculate the answers to each part of the question.