(a) What is the speed of the combined blocks after the collision? (b) What angle

Question

                    (a) What is the speed of the combined blocks after the collision?

(b) What angle does the velocity of the combined blocks make with the +x-axis after the collision?

(c) What is the magnitude of the decrease in kinetic energy of the system of two blocks due to the collision?

Two blocks are sliding on a horizontal frictionless surface with velocities shown in the sketch. Block A has mass 0.500 kg and block B has mass 0.200 kg. The two blocks have a perfectly inelastic collision and stick together after the collision. What is the speed of the combined blocks after the collision? What angle does the velocity of the combined blocks make with the +x-axis after the collision? What is the magnitude of the decrease in kinetic energy of the system of two blocks due to the collision?

Explanation / Answer

(A) since the collision is perfectly inelastic momentum of the combined system will bremail conserved

so initial momentum = 0.200*0.500 in x direction + 0.200*0.400 in y direction = 0.10 i^ + 0.08 j^

so magnitude of initial momentum = 0.128 kg m/s

so after collision combined speed = total momentum/combined mass = 0.128/ (0.200+ 0.500) = 0.183 m/s.

(B) the angle(a) with +ve x-axis will be given by , tan a = 0.08/0.10 = 0.8 , (because tan a is equal to momentum in y direction divided by momentum in x direction , where a is the angle made by resultant momentum witth x-axis)

  

so a = tan-1(0.8) = 38.66 degrees.

(C) initial kinetic energy = 1/2 (m1v1^2 + m2v2^2) = 1/2(0.500*0.200^2 + 0.200*0.400^2) = 0.026 J

final kinetic energy =1/2 ((m1+m2 ) *0.183^2) = 1/2 (0.700*0.183^2) = 0.0117 or 0.012 J

so magnitude of the decrease in kinetic energy =initial ke - final ke = 0.026 - 0.0117 = 0.0143 J . ans

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

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