A basketball explodes into three separate pieces. Two pieces of equal mass shoot off at 20.3 m/s, one travelling to the south and one to the west, perpendicular to each other. The third piece has a mass 4 times the size of one of the other pieces. What is the speed of the third piece? (please give me answer and show how to do it thank you)
Final momentum=zero (same as initial momentum)
in the south direction:
M*S+4M*?=0 S=20.3 South
South compoent= 1/4 *20.3 in N direction.
East direction:
M*(-20.4E)+4M(?)=0
East component=20.4/4
Speed=sqrt(S + E)=20.3*sqrt(1/16+1/16)
= 20.3/4 sqrt2
To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum of an isolated system remains constant before and after a collision.
Let's denote the mass of one of the smaller pieces as m. Since there are two equal mass pieces shooting off, the total mass of these two pieces is 2m. According to conservation of momentum, the total momentum before the explosion is equal to the total momentum after the explosion.
Before the explosion:
Total momentum = (mass of first piece) × (velocity of first piece) + (mass of second piece) × (velocity of second piece)
After the explosion:
Total momentum = (mass of third piece) × (velocity of third piece)
We can write this as an equation:
2m × 20.3 m/s = (4m) × (velocity of third piece)
Simplifying the equation:
40.6 m/s = 4m × (velocity of third piece)
Dividing both sides by 4m:
10.15 m/s = velocity of third piece
Therefore, the speed of the third piece is 10.15 m/s.
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the explosion is equal to the total momentum after the explosion.
Let's assign variables to represent the masses and velocities of the pieces:
Let the mass of each of the two smaller pieces be 'm', and the velocity of one piece (shooting off to the south) be 'v1' and the velocity of the other piece (shooting off to the west) be 'v2'.
Given that the two smaller pieces shoot off with equal mass and velocities of 20.3 m/s in perpendicular directions, we have:
(m * v1) + (m * v2) = total momentum before explosion (Equation 1)
Now, let the mass of the third piece be '4m', and its velocity be 'v3'. According to the principle of conservation of momentum:
(m * v1) + (m * v2) + (4m * v3) = total momentum after explosion (Equation 2)
We can now solve for 'v3'. But first, let's substitute the given values into Equation 1 to find the total momentum before the explosion:
(2m) * (20.3 m/s) + (2m) * (20.3 m/s) = 2m * (20.3 m/s + 20.3 m/s) = 2m * (40.6 m/s) = 81.2 m * s
Now we can use this value in Equation 2 to solve for 'v3':
81.2 m * s + (4m * v3) = (6m * v3)
Rearranging the equation:
(6m * v3) - (4m * v3) = 81.2 m * s
2m * v3 = 81.2 m * s
Dividing both sides of the equation by 2m:
v3 = 81.2 m * s / (2m) = 40.6 m/s
Therefore, the speed of the third piece is 40.6 m/s.