A 2900 kg truck traveling north at38 km/h turns east and accelerates to58 km/h.

Question

A 2900 kg truck traveling north at38 km/h turns east and accelerates to58 km/h. (a) What is the change in the truck's kineticenergy?
J

(b) What is the magnitude of the change in the linear momentum ofthe truck?
kg·m/s

(c) What is the direction of the change in the linear momentum ofthe truck?
° (measured clockwise from east)
(a) What is the change in the truck's kineticenergy?
J

(b) What is the magnitude of the change in the linear momentum ofthe truck?
kg·m/s

(c) What is the direction of the change in the linear momentum ofthe truck?
° (measured clockwise from east)

Explanation / Answer

v1 = 38000 / 3600 = 10.56 m/s = vy v2 = 58000 / 3600 = 16.11 m/s = vx KE = 1/2 m (v22 -v12) = 1/2 * 2900 * (259 - 111) = 2.14 * 10E5J p2 - p1 =p2 + (-p1)    the vector"change" in momentum p1 = m v1 = 3.06 * 10E4 kg-m/s p2 = m v2 = 4.67 * 10E4 kg-m/s p = (3.062 + 4.672) * 10E4 = 5.58* 10E4 kg-m/s    magnitude of change inmomentum tan = py / px = -10.56 / 16.11 =-.655     = -33.2 deg = 33.2 deg Sof E KE = 1/2 m (v22 -v12) = 1/2 * 2900 * (259 - 111) = 2.14 * 10E5J p2 - p1 =p2 + (-p1)    the vector"change" in momentum p1 = m v1 = 3.06 * 10E4 kg-m/s p2 = m v2 = 4.67 * 10E4 kg-m/s p = (3.062 + 4.672) * 10E4 = 5.58* 10E4 kg-m/s    magnitude of change inmomentum tan = py / px = -10.56 / 16.11 =-.655     = -33.2 deg = 33.2 deg Sof E

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.