A block of mass 1.8 kg slides down a 30 degree incline from a height of 3.6m. Wh

Question

A block of mass 1.8 kg slides down a 30 degree incline from a height of 3.6m. When it reaches the bottom it collides elastically with a mass 6.4 kg. There's no friction A what would the speed of the 1.4 kg be at the bottom? B. After collision what will the final velocities of the two object be? A block of mass 1.8 kg slides down a 30 degree incline from a height of 3.6m. When it reaches the bottom it collides elastically with a mass 6.4 kg. There's no friction A what would the speed of the 1.4 kg be at the bottom? B. After collision what will the final velocities of the two object be?

Explanation / Answer

er:  At the bottom of the slope, before striking the heavier mass, the small mass has velocity Vo

Vo is such that

1/2mVo^2 = mgh
Vo = sqrt(2*gh) = sqrt(2*9.8*3.6)
Vo = 8.4m/s

Before and after the shock, we have

a) Conservation of momentum

mVo = mV1 + MV2
18.48 = 1.8*V1 + 7*V2
V1 = (18.48 - 7V2)/1.8
V1 = 8.4 - 3.18V2


b) Conservation of energy

1/2m*Vo^2 = 1/2mV1^2 + 1/2 MV2^2
77.616 = 1.1*V1^2 + 3.5V2^2
77.616 = 1.1*(8.4 - 3.18V2)^2 + 3.5V2^2

Solving for V2 gives

V2 = 4.019m/s
V1 = -4.379m/s

b) With such a velocity the smaller mass has kinetic energy of

Ek = 1/2 m*V2^2
Ek = 1/2*1.8*(-4.379m/s)^2
Ek = 21.09 J

This will be transformed into potential energy as it goes up the ramp to height h

mgh = 21.09J
h = 21.09/(9.8*1.8) = 0.978 m

At an angle of 30 degrees this represent a distance on the incline of

d = h/sin(30) = 1.95 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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