Answer to Question #167195 in Mechanics | Relativity for Gabriela

Question

a truck P (900kg)starts from rest runs down a frictionless slope from point a to b which is 101.4 meters, from a height at point a of 33.8 meters to b 0 meters,truck P collided with truck Q (337,5kg)at point b and continues to move with a speed of 19,5 m.s¹.,what is tuck Q velocity after collision

Answer 1

We have,

Mass of truck P, $m_1$ = 900 kg

Mass of truck Q, $m_2$ = 337.5kg

As the initial velocity of truck P = 0 m/s

Using the third equation of motion,

$v^2 - u^2 = 2\times gcos\theta\times s$

$v^2 = 2\times 10 \times 0.33\times 101.4$

$v = 25.86$ m/s

Now using the conservation of momentum,

$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$ ( Here $u_1 = v$ )

$900\times 25.86 + 0 = 900\times 19.5 + 337.5\times v_2$

$v_2 = \dfrac {5724}{337.5}$

$v_2 = 16.96 m/s$

Hence , the velocity of truck Q after collision is $16.96$ m/s