What Is the Velocity of the Third Piece After the Ball Explodes?

[DrVirus]

this problem needs alittle understanding about momentum, something i don't understand that much.

A ball of mass 600g is moving 12 m/s[e]. The ball suddenly explodes into three equal mass pieces. One piece heads north at 6m/s, the second piece heads west at 5m/s. Determine the velocity of the third piece.

Can someone please show the calculations for me please, thnx a lot.

 

[HallsofIvy]

Pretty straight forward. The initial velocity is <12, 0> (I'm taking my axes to point east and north). The intitial momentum vector is (mass times velocity) 0.6 KG <12 m/s, 0> = <7.2, 0>.

After the "explosion", each piece has mass 200 mg= 0.2 kg. The piece heading north at 6 m/s has momentum .2<0, 6>= <0, 1.2>. The piece heading west at 5 m/s has momentum .2<-0.5, 0>= <-0.1, 0>.
If we let <x, y> be the velocity of the third piece, its momentum is
<.2x, .2y>.

Conservation of momentum says: <7.2, 0>= <0- 0.1+ 0.2x, 1.2+ 0+ .2y> or 0.2x- .1= 7.2 and 0.2y+ 1.2= 0.

 

[M98Ranger]

Momentum is a fundamental concept in physics that describes the quantity of motion an object possesses. It is a vector quantity, meaning it has both magnitude and direction. In this problem, we are dealing with a situation where a ball with a certain momentum suddenly explodes into three equal mass pieces, each with their own momentum. To find the momentum of each piece, we need to use the formula p = mv, where p is momentum, m is mass, and v is velocity.

First, we need to find the total momentum of the original ball before it exploded. This can be done by multiplying the mass (600g or 0.6kg) by the initial velocity (12m/s).

p = mv
p = (0.6kg)(12m/s)
p = 7.2 kg m/s

Since momentum is conserved in a closed system, the total momentum before the explosion will be equal to the total momentum after the explosion. This means that the sum of the individual momentums of the three pieces will also be equal to 7.2 kg m/s.

Now, we can use this information to find the velocity of the third piece. Since we know the velocities of the other two pieces (6m/s north and 5m/s west), we can use the Pythagorean theorem to find the magnitude of the third piece's velocity.

v^2 = (6m/s)^2 + (5m/s)^2
v^2 = 36m^2/s^2 + 25m^2/s^2
v^2 = 61m^2/s^2
v = √61m/s ≈ 7.81m/s

To find the direction of the third piece's velocity, we can use trigonometry. Since the first two pieces are heading north and west, the third piece must be heading in the direction of the hypotenuse of a right triangle with sides of 6m/s and 5m/s. This angle can be found using the inverse tangent function.

θ = tan^-1(5m/s ÷ 6m/s)
θ ≈ 39.8°

Therefore, the velocity of the third piece is approximately 7.81m/s in a direction of 39.8° from the positive x-axis.

I hope this helps in understanding momentum better and how it applies to this specific problem. Remember, momentum is always conserved in a closed