In the figure, a 2.1 kg box of running shoes slides on a horizontal frictionless

Question

In the figure, a 2.1 kg box of running shoes slides on a horizontal frictionless table and collides with a 1.5 kg box of ballet slippers initially at rest on the edge of the table, at height h = 0.73 m. The speed of the 2.1 kg box is 3.0 m/s just before the collision. If the two boxes stick together because of packing tape on their sides, what is the magnitude of their velocity just before they strike the floor? Use 2 different methods and see if you get the same result (use Conservation of Energy Or Kinematics&Dynamics; for the projectile motion part).

Explanation / Answer

2) for collision, Using momentum conservation,

initial momentum = final momentum

2.1 x 3 + 1.5x0 = (2.1+ 1.5) v

v = 1.75 m/s

after collision both blocks move with 1.75 m/s together.

Using conservation of energy,

initial PE + KE = final PE + KE

mgh + mv^2 /2 = 0 + m vf^2 /2

(m x9.8 x 0.73) + (m x 1.75^2 /2 ) = m vf^2 /2

vf = 4.17 m/s

Using Kinetmatic and DYnamics,

initial velocity was all in horizontal.

so initially vertical velocity was zero. and it have to go down 0.73 under the effect of gravity.

using vyf^2 - Vyi^2 = 2(a)(y)

vyf^2 - 0 = 2(-9.8)(-0.73)

vfy = 3.78 m/s

vfx = v = 1.75 m/s

vf = sqrt(vfx^2 + vfy^2) = 4.17 m/s

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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