Alevel physics. momentum questions.

kinematic equation used, g=10m/s/s approx

(a) 39.6 m/s and 30.7 m/s

Explanation:

Use the formula for speed as a function of distance made in a uniformly decelerated motion:

with v_s the instantaneous speed at distance s and v_0 the initial speed (right after the explosion). "a" is the acceleration due to friction force, with negative sign in front of that term reflecting the fact the friction force acts against the direction of the motion. The scene after the explosion implies the fragments have come to a halt with the respective distances shown in the figure, i.e., for each fragment:

and the initial speed v_0 remains to be determined:

The deceleration "a" due to friction can be found using the information we are given: the mass of a fragment and the coefficient of dynamic friction of 0.4:

So the initial velocities just after the explosion, as implied by the distances of 200m (v01) and 120m (v02) are, respectively:

(b) The speed of the third fragment is 31.7 m/s

Explanation:

Use the law of conservation of the momentum. At the time of the explosion there were three fragments. For two of them we have determined the initial speed in (a). Now we know that the total momentum of the system (container) right before the fragments were set into motion was 0. The total of the moment vectors (magnitudes with their directions) should still be 0 right after the explosion. Given the angle between the paths of fragment 0.5kg and 1kg, the total vector of their momentum can be calculated

and from the conservation law we know that the momentum of the third piece must be

in particular, its magnitude will be same as the magnitude of the resultant vector, counteracting (at an angle 180 degrees from the resultant). This will eventually allow us to determine the speed of the third fragment. The magnitudes are:

and the resulting moment:

and so the speed of the third fragment is 

(c) The total kinetic energy is 1617 Joules

Explanation:

(d) The area will be proportional to the fourth power of the initial velocities.

Explanation:

Consider the area of a circle with radius equal to the distance a fragment traveled. We can choose a fragment with largest such distance or choose the average of the areas for each fragment, or another geometric measure, however, this choice won't affect the qualitative answer.

The distance of a fragment "i"  traveled as a function of the initial speed is

The circular area is then

The area due to a fragment with initial velocity is proportional to the fourth power of that velocity. This can be generalized to all fragments by assuming a common factor amplifying the velocities. Such factor will also show up in the fourth power in the area formula above, justifying the the effect of an amplification of the initial speed has a fourth-power effect on the area of spread. 

I will answer in English.

Ok, we know that a ball is trowed up, and the maximum height is 21.4m

If we suppose that the initial height of the ball is h = 0 (this means that is trowed from the ground) the equations of movement will be:

for the acceleration we only have the gravitational acceleration, so we have:

a(t) = -9.8m/s^2

for the velocity we can integrate over time and get:

v(t) =  (-9.8m/s^2)*t + v0

where v0 is the initial speed.

For the position we integrate over the time again, and as the initial position is 0m, here we do not have any integration constant.

p(t) = (1/2)(-9.8m/s^2)*t^2 + v0*t

a) the maximum height is reached when the velocity is equal to zero, this happens at the time:

(-9.8m/s^2)*t + v0 = 0

t = v0/(9.8m/s^2)

now we can replace this time in the equation for the position:

p = 21.4m = (-4.9m/s^2)*(v0/(9.8m/s^2))^2 + v0^2/(9.8m/s^2)

and solve it for v0, i will stop writing the units so it is easier to read:

21.4 = -4.9*(v0/9.8)^2 + v0^2/9.8

21.4 = v0^2*(0.05)

v0 = √(21.4/0.05) = 20.7

so the initial speed is 20.7 m/s.

b) the things i supposed are:

The initial position is p(0) = 0

there are no things like air resistance or wind.

c) we can take t = 3.05s and put this in the position equation:

p(3.05s) =  (1/2)(-9.8m/s^2)*(3.05s)^2 + 20.7m/s*3.05s = 17.5m

3.05 seconds after the launch, the ball will be 17.5 m above the ground.

d) we put t = 3.05s in the velocity equation:

v(3.05s) =  (-9.8m/s^2)*3.05s + 20.7m/s = -9.19m/s

which means that now the ball is falling down.