Question

A 1240 kg car traveling south at 24 m/s with respect to the ground collides with and attaches to a 2510 kg delivery truck traveling east at 16 m/s.

A)Determine the direction of the velocity of the two vehicles when locked together just after the collision. Take east direction as zero angle.

B)Determine the magnitude of the velocity of the two vehicles when locked together just after the collision.

Answer 1

"Let's take m_c: - mass of the car m_t: - mass of the truck V_c vector: - velocity of car before collission V_t vector: - velocity of truck before collission i unit vector along East j unit vector along North According to the law of conservation of linear mmomentum, Momentum before collission = Momentum after collission P before Momentum before collission = m_cV_c vector + m_cV_t vector m_c = 1240 kg v_c vector = -24 j m/s m_c = 2510 kg V_t vector = 16 im/s so P before = 1240 times -24 j + 2510 times 16 i = -29780 j + 40160 i Hence angle mode by the Pbefore vector with East direction is given by \r\n theta = tan^-1 (29760/40160) = 36.54 in IV^thQuadrant so that is the direction of final velocity.\r\n B) After the collission both the vehicles stick together. so the net mass after the collission is given by m = m_c + m_t = 3750 kg V vector = velocity after collission so momentum after collission is m_v vector. m_v vector = 40160(i) + 29760(-j) so, V vector = 40160(i) + 29760(-j) /m = 40160/3750 (i) - 29760/3750 (j) so magnitude of final velocity