(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:
[tex]v_s^2 = v_0^2 -2sa[/tex]
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:
[tex]0 = v_0^2 -2sa[/tex]
and the initial speed v_0 remains to be determined:
[tex]v_0=\sqrt{2sa}[/tex]
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:
[tex]F_{friction} = \mu\cdot m\cdot g \implies a = \mu\cdot g = 0.4\cdot 9.8\frac{m}{s^2}=3.92\frac{m}{s^2}[/tex]
So the initial velocities just after the explosion, as implied by the distances of 200m (v01) and 120m (v02) are, respectively:
[tex]v_{01}=\sqrt{2\cdot200m\cdot 3.92\frac{m}{s^2}}=39.6\frac{m}{s}\\v_{02}=\sqrt{2\cdot120m\cdot 3.92\frac{m}{s^2}}=30.7\frac{m}{s}[/tex]
(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
[tex]\overline{p}_{12} = m_1 \overline{v}_1 +m_2\overline{v}_2\\[/tex]
and from the conservation law we know that the momentum of the third piece must be
[tex]\overline{p}_3=-\overline{p}_{12}[/tex]
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:
[tex] |\overline{p}_{1}| = 0.5kg\cdot 39.6 \frac{m}{s} = 19.8 kg\frac{m}{s}\\|\overline{p}_{2}|=1.0kg\cdot 30.7\frac{m}{s}=30.7kg\frac{m}{s}[/tex]
and the resulting moment:
[tex] |\overline{p}_{12}| = \sqrt{|\overline{p}_1|^2+|\overline{p}_2|^2+2|\overline{p}_1||\overline{p}_2|\cos 40^\circ}=47.6 kg\frac{m}{s}=|\overline{p}_3|[/tex]
and so the speed of the third fragment is
[tex]v_{03} = \frac{|\overline{p}_3|}{1.5kg}=\frac{47.6kg\frac{m}{s}}{1.5kg}=31.7\frac{m}{s}[/tex]
(c) The total kinetic energy is 1617 Joules
Explanation:
[tex]E_k = \frac{1}{2}\sum_{i=1}^3 m_i v_{0i}^2 = 0.5\cdot(0.5\cdot 39.6^2+1\cdot 30.7^2+1.5\cdot 31.7^2)kg\frac{m^2}{s^2}=1617.0J[/tex]
(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
[tex]s_i = \frac{v_{0i}^2}{2a}[/tex]
The circular area is then
[tex]A_i = \pi(\frac{v_{0i}^2}{2a})^2\implies A_i \propto v_{0i}^4[/tex]
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 answer: the effect of an amplification of the initial speed has a fourth-power effect on the area of spread.
