A block with mass m 1 = 0.450 kg is released from rest on a frictionless track a

Question

A block with mass

m1 = 0.450 kg

is released from rest on a frictionless track at a distance

h1 = 3.10 m

above the top of a table. It then collides elastically with an object having mass

m2 = 0.900 kg

that is initially at rest on the table, as shown in the figure below.

(a) Determine the velocities of the two objects just after the collision. (Enter the magnitude of the velocity.)

(b) How high up the track does the 0.450-kg object travel back after the collision?
m

(c) How far away from the bottom of the table does the 0.900-kg object land, given that the height of the table is

h2 = 1.95 m?

m

(d) How far away from the bottom of the table does the 0.450-kg object eventually land?
m

Explanation / Answer

Here ,

a) m1 = 0.450 Kg

m2 = 0.90 Kg

for the velocity before collision

u1 = sqrt(2 * g * 3.1) = 7.8 m/s

Now , after the collision

0.450 * v1 + 0.90 * v2 = 0.450 * 7.8 -----(1)

as the collision is elastic

v2 - v1 = 7.8 ----(2)

solving 1 and 2

v1 = -2.6 m/s

v2 = 5.2 m/s

for the magnitude

v1 = 2.6 m/s

v2 = 5.2 m/s

b)

height of 0.450 Kg block= v1^2/(2 * g)

height of 0.450 Kg block= 2.6^2/(2 * 9.8)

height of 0.450 Kg block= 0.345 m

c)

time of flight , t = sqrt(2 * 1.95/9.8)

t = 0.631 s

distance from the table = 0.631 * 5.2 m

distance from the table = 3.28 m

d)

eventually ,

as the speed of block will again be 2.6 m/s

distance of landing = 2.6 * 0.631

distance of landing = 1.64 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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