A m1 = 0.550 kg block is released from rest at the top of a frictionless track h

Question

A m1 = 0.550 kg block is released from rest at the top of a frictionless track h1 = 3.00 m above the top of a table. It then collides elastically with a 1.00 kg block that is initially at rest on the table, as shown in Figure P6.57. Figure P6.57 (a) Determine the velocities of the two objects just after the collision. velocity of m1 m/s velocity of m2 m/s (b) How high up the track does the 0.550 kg object travel back after the collision? m (c) How far away from the bottom of the table does the 1.00 kg object land, given that the table is 1.90 m high? m (d) How far away from the bottom of the table does the 0.550 kg object eventually land? m

Explanation / Answer

m1 = 0.550 kg
h1 = 3.00 m above table
m2 = 1.00 kg
h2 = 1.90 m from ground

(a) In this scenario, m1 slides down and hits m2 from the side. Relative to the table, it converts all its gravitational potential energy into kinetic energy.

PE = KE
mgh1 = (1/2)mv2

gh1 = (1/2)v2

(9.81 m/s2)(3.00 m) = (1/2)v2

v = (9.81 * 3 * 2) m/s
v = 7.67 m/s

if there's an elastic collision, then no momentum is lost and energy is conserved:

m1*v = m1*v1 + m2*v2
(1/2)m1*v2  = (1/2)m1*v12 + (1/2)m2*v22

We have two equations and two variables (v1 and v2), so we solve:
0.55*7.67 = 0.55*v1 + 1*v2 = 0.55*v1 + v2
4.22 = 0.55*v1 + v2
v2 = 4.22 - 0.55*v1

0.5*0.55*(7.67)2 = 0.5*0.55*v12 + 0.5*1*v22
16.2 = 0.28*v12 + 0.5*v22
32.4 = 0.56*v12 + v22

32.4 = 0.56*v12 + (4.22 - 0.55*v1)2
32.4 = 0.56*v12 + 17.81 - 4.64v1 + 0.31*v12

0 = -14.55 - 4.64*v1 + 0.87*v12

v1 = 7.55 or -2.22

Since v1 applies to m1, it must travel left, which is the negative direction, so

v1 = -2.22 m/s

v2 = 5.44 m/s

(b)

Now, all the kinetic energy of m1 is being converted into potential energy

(1/2)m1*v12 = m1*g*h

(1/2)*v12 = g*h

h = (1/2)*v12/g = 0.5*(-2.22)2/9.81

h = 0.251 m

(c)

We have the initial velocity and we know how far it drops. We need to figure out how long it would take to free fall to the bottom. We use this time to calculate how far it travels.

y = h2 - 0.5*g*t2

0 = 1.90 - 0.5*9.81*t2

t = 0.622 s

The initial velocity in the x-direction is 5.44 m/s. So, if it hits the ground in 0.622 s, it travels:

x = v2*t

x = 5.44*0.622

x = 3.38 m

(d)

Now we have to figure out the velocity of m1 when it gets to the edge of the table by equating its potential energy to its kinetic energy.

m1*g*0.251 = 0.5*m1*v32

9.81*0.251 = 0.5 * v32

v3 = 2.22 m/s

It'll also hit the ground in 0.622 seconds.

x = v3*t

x = 2.22 * 0.622

x = 1.38 m

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