Question

A 732 kg car stopped at an intersection is rear-ended by a 1720 kg truck moving with a speed of 15.5m/s. If the car was in neutral and its breaks were off, so the collisions are approximately elastic, find the final speed of both vehicles after the collisions.

Answer 1

Answer:

Truck= 6.25 m/s

Car= 21.7 m/s

Explanation:

For elastic collision, the truck's final velocity is given by

${v_1} = \frac{{\left( {{m_1} - {m_2}} \right)}}{{\left( {{m_1} + {m_2}} \right)}}{u_1}$

The car's final velocity is given by

${v_2} = \frac{{2{m_1}}}{{\left( {{m_1} + {m_2}} \right)}}{u_1}$

Where m and u denote masses and velocity respectively, subscripts 1 and 2 denote truck and car respectively.

Substituting 1720 Kg for mass of truck, 732 for mass of car and initial velocity as 15.5 m/s

$v_1=\frac{1720-732}{1720+732}\times 15.5=6.25 m/s$

$v_2=\frac {2\times 1720}{1720+732}\times 15.5=21.7 m/s$

Therefore, the final velocity of truck is 6.25 m/s while the car's final velocity is 21.7 m/s

Answer 2

The final velocity of the car is 6.22 m/s and the final velocity of the truck is 21.72 m/s.

The given parameters;

• mass of the car, m = 732 kg
• mass of truck, M = 1720 kg
• initial velocity of the truck, u₂ = 15.5 m/s

The final velocity of both vehicles is calculated by applying the principle of conservation of linear momentum;

$m_1u_1 \ + Mu_2 = m_1v_1 + Mv_2\\\\732(0) + 1720(-15.5) = -732v_1 \ + \ 1720v_2\\\\ -26,660 = -732v_1 + 1720v_2$

Apply one-directional linear velocity equation;

$u_1 + v_1 = u_2 + v_2\\\\0 - v_1 = -15.5 + v_2\\\\v_1 = 15.5 - v_2$

substitute the value of v₁ into the above equation;

$-26,600 = -732(15.5 - v_2) \ + 1720v_2\\\\ - 26,600 = - 11346 + 732v_2 + 1720v_2\\\\ -15,254 = 2,452v_2 \\\\ v_2 = \frac{-15,254}{2,452} \\\\v_2 = - 6.22 \ m/s\\\\v_1 = 15.5 + 6.22 = 21.72 \ m/s$

Thus, the final velocity of the car is 6.22 m/s and the final velocity of the truck is 21.72 m/s.

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