3. A car travelling at 12 m/s into a stationary truck of about 10 times the cars mass. a. If the collision was completely inelastic, what velocity would the two travel at if the stuck together? b. If the collision was completely elastic, what would be the velocities of the car and truck after the collision? c. In order to exert a force of only 3500N on the truck during the collision, how much time would the collision have to take?

(b) For the elastic collision, the final velocity of the car is 9.81 m/s backwards and the final velocity of the truck is 2.19 m/s forward.

(c) The time taken to exert the given force is 0.00625 m (s).

The given parameters;

Initial velocity of the car, u₁ = 12 m/s
Initial velocity of the truck, u₂ = 0
Mass of the car, = m
Mass of the truck, = 10m

(a) The final velocity of the two vehicles if the collision was inelastic is calculated as follows;

[tex]m_1 u_1 + m_2u_2 = v(m_1+ m_2)\\\\12m + 10m(0) = v(m + 10m)\\\\12m = v(11m)\\\\v = \frac{12m}{11m} \\\\v = 1.1 \ m/s[/tex]

(b) The final velocity of the two vehicles if the collision was elastic is calculated as follows;

[tex]m_1 u_1 + m_2u_2 = m_1v_1 + m_2v_2\\\\\12m \ + \ 10m(0) = mv_1 + 10mv_2\\\\12m = m(v_1 + 10v_2)\\\\12 = v_1 + 10 v_2\ \ - --(1)[/tex]

Apply one-directional velocity equation:

[tex]u_1 +v_1 = u_2 + v_2\\\\12 + v_1 = 0 + v_2\\\\12+ v_1 = v_2 \ \ --- (2)[/tex]

Substitute the value of [tex]v_2[/tex] into equation (1);

[tex]12 = v_1 + 10(12 + v_1)\\\\12= v_1 + 120 + 10v_1\\\\12- 120 = 11v_1\\\\-108 = 11v_1\\\\v_1 = \frac{-108}{11} \\\\v_1 = -9.81 \ m/s\\\\[/tex]

Solve for [tex]v_2[/tex];

[tex]v_2 = 12 + v_1\\\\v_2 = 12 - 9.81\\\\v_2 = 2.19 \ m/s[/tex]

Thus, for the elastic collision, the final velocity of the car is 9.81 m/s backwards and the final velocity of the truck is 2.19 m/s forward.

(c)

The change in the momentum of the truck is calculated as;

[tex]\Delta P = m_2(v_2 - u_2)\\\\\Delta P = 10m(2.19)\\\\\Delta P = 21.9m[/tex]

The time taken to exert the given force is calculated as follows;

[tex]Ft = \Delta P\\\\t = \frac{\Delta P}{F} \\\\t = \frac{21.9 \ m}{3500} \\\\t = 0.00625 \ m (seconds)[/tex]

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