I know that the momentum of collision before= after collision momentum

so according to that (m1xv1)+(m2xv2)=(m1+m2)v
how do i do this

v2 = 4.68*10^9 [kg]

Explanation:

In this problem we must show that the speed the mass of the asteroid is approximately equal to that mentioned in numeral a).

We need to identify the initial data.

m1 = 300 [kg] (mass of the dart)

v1 = 6250 [m/s]

v2 = 0.4 [mm/s] = 0.0004[m/s]

Using the equation for this type of collision

(m1*v1) + (m2*v2) = (m1+m2)*v2

Now the key to solving this problem is the term relative speed, relative speed is defined as the speed of one body relative to the other in a given instant. For this case it has a relative speed of 6250 [m/s] of the Dart with respect to the asteroid before impact, so the speed of the asteroid before impact is taken as zero.

Therefore:

(m1*v1) = (m1+m2)*v2

(300*6250) = (300*0.0004) + 0.0004*v2

v2 = 4.68*10^9 [kg]

This is approximately equal to the value of point a.

none, its a machine

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~