A very large firecracker explodes in the hallway into 3 fragments. Physics students rush to the scene to enrich their knowledge of the law of conservation of momentum. They report that
Fragment 1: mass=1.23kg, v=8.70m/s, north
Fragment 2: mass =2.60kg,v=6.10m/s,east
fragment 3: ?????
A) calculate the magnitude and direction of the momentum of the 3 fragment. (Keep in mind that momentum is a vector quantity)
b) if the 3rd fragment has a speed of 11.6m/s what is its mass?
Please show all work I would greatly appreciate it Bc I am extemtly lost
To calculate the momentum of each fragment, we need to use the formula for momentum:
Momentum (p) = mass (m) * velocity (v)
For Fragment 1:
Mass (m1) = 1.23 kg
Velocity (v1) = 8.70 m/s (north)
So, momentum of Fragment 1 (p1) = m1 * v1
For Fragment 2:
Mass (m2) = 2.60 kg
Velocity (v2) = 6.10 m/s (east)
So, momentum of Fragment 2 (p2) = m2 * v2
Now, let's calculate the magnitudes and directions of the momenta:
Magnitude of momentum (p) is given by the formula:
p = sqrt(p_x^2 + p_y^2)
Direction (θ) of momentum is given by the formula:
θ = atan(p_y/p_x)
For Fragment 1:
p1 = sqrt((m1 * v1)^2)
θ1 = atan(v1/0)
For Fragment 2:
p2 = sqrt((m2 * v2)^2)
θ2 = atan(0/v2)
Now, let's find the values:
For Fragment 1:
p1 = sqrt((1.23 kg * 8.70 m/s)^2)
= sqrt(8.49 kg*m/s)^2
= sqrt(72.1203 kg^2*m^2/s^2)
≈ 8.49 kg*m/s (north)
Direction (θ1) of p1 is north, as given.
For Fragment 2:
p2 = sqrt((2.60 kg * 6.10 m/s)^2)
= sqrt(15.86 kg*m/s)^2
= sqrt(251.796 kg^2*m^2/s^2)
≈ 15.86 kg*m/s (east)
Direction (θ2) of p2 is east, as given.
Now, for Fragment 3, if its speed (v3) is 11.6 m/s, we can calculate its mass (m3) using the equation:
p3 = m3 * v3
Since we know the magnitude of momentum (p3) and speed (v3), we can substitute these values into the equation and solve for m3:
p3 = m3 * v3
m3 = p3 / v3
Given that p3 = 15.86 kg*m/s (calculated for Fragment 2)
m3 = 15.86 kg*m/s / 11.6 m/s
≈ 1.37 kg
Therefore, the mass (m3) of Fragment 3 is approximately 1.37 kg.
To solve this problem, we can use the law of conservation of momentum, which states that the total momentum before an event is equal to the total momentum after the event, as long as there are no external forces acting on the system. In this case, the explosion of the firecracker is the event.
a) To calculate the momentum of each fragment, we can use the formula:
momentum = mass * velocity
For fragment 1:
mass = 1.23 kg
velocity = 8.70 m/s
momentum1 = 1.23 kg * 8.70 m/s = 10.731 kg·m/s (north)
For fragment 2:
mass = 2.60 kg
velocity = 6.10 m/s
momentum2 = 2.60 kg * 6.10 m/s = 15.86 kg·m/s (east)
Now, since momentum is a vector quantity, we need to find the resultant momentum of fragments 1 and 2. This can be done by using vector addition.
To find the magnitude and direction of the resultant momentum, we can use the Pythagorean theorem and trigonometric functions:
resultant momentum = sqrt((momentum1)^2 + (momentum2)^2)
magnitude = sqrt((10.731 kg·m/s)^2 + (15.86 kg·m/s)^2) = 19.041 kg·m/s
To find the direction, we can use the inverse tangent function:
direction = atan(momentum2 / momentum1)
= atan(15.86 kg·m/s / 10.731 kg·m/s)
≈ 1.057 radians (east of north)
Therefore, the magnitude of the momentum is 19.041 kg·m/s, and the direction is approximately 1.057 radians east of north.
b) If the third fragment has a speed of 11.6 m/s, we can use the momentum formula to find its mass:
momentum3 = mass3 * velocity3
Given momentum3 = 19.041 kg·m/s (from the previous calculation) and velocity3 = 11.6 m/s, we can rearrange the formula as follows:
mass3 = momentum3 / velocity3
mass3 = 19.041 kg·m/s / 11.6 m/s ≈ 1.642 kg
Therefore, the mass of the third fragment is approximately 1.642 kg.