A block with mass m1 = 0.300 kg moving on a frictionless surface at an initial s

Question

A block with mass m1 = 0.300 kg moving on a frictionless surface at an initial speed of v1i = 2.50 m/s undergoes an elastic collision with a block of unknown mass m2 moving in the same direction as the first block at an initial speed of v2i = 1.12 m/s. After the collision, the first cart continues in its original direction at v1f = 0.280 m/s.

(a) What is the mass of the second block m2?

(b) What is the speed of the second block v2f after impact?

***The following is not the correct solution...***

For elastic collision wehave 0.3x 2.50 +m, ×1.12-m,× 0.280 +m, 0.75 +1.12m, = 0.280x 0.3 +m,½ 0.75-0.084-1.12m +mvi 0.666-m2(x%-1.12) 0.666 ( -1.12) Kineticenergy conservation 0.5x 0.3x (2.5)2 +0.5× m, x (1.12)-0.5x 0.3x (0.280)2 +05xmw2 0.937 + 0.6272m = 0.01 176 + 0.5x m y? 272 0.925 = 0.5 x m2y2-0.6272m, 0.925-m, (0.5 925m(0.5v2-0.6272) -0.6272) 0.925 0.5 2-0.6272) 0.925 (05W2-06272 0.666 (*-112) (0.5v,-0.6272) (4-112) 0.3330.418-0.925v2-1.036 -0592u? =-0.618 2_0.618 20.592 1.02m / s Substituting this in (1)

Explanation / Answer

Using the equation for elastic collision as

V2f - V1f = V1i - V2i

V2f - 0.28 = 2.5 - 1.12

V2f = 1.66 m/s                   eq-1

Using conservation of momentum

m1 V1i + m2 v2i = m1 V1f + m2 V2f

0.3 x 2.5 + 1.12 m2 = 0.3 x 0.28 + m2 (1.66)

m2 = 1.23 kg

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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