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

Question

A m1 = 0.550 kg block is released from rest at the top of a frictionless track h1 = 2.65 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.

http://www.webassign.net/sf/p6_57alt.gif

(a) Determine the velocities of the two objects just after the collision.
velocity of m1:
velocity of m2:

(b) How high up the track does the 0.550 kg object travel back after the collision?

(c) How far away from the bottom of the table does the 1.00 kg object land, given that the table is 2.20 m high?

(d) How far away from the bottom of the table does the 0.550 kg object eventually land?

Answer 1

Nice drawing!

a) The velocity of m1 just before impact is found from
Pe=Ke
m1gh=(1/2) m1 U1^2
U1=sqrt(2 gh)=7.2m/s

Conservation of momentum
m1U1 +m2U2=m1V1 + m2V2
It helps to know that U2= 0

m1U1 = m1V1 + m2V2
also
What about conservation of kinetic energy? Ke=(1/2) mV^2
Ke1+Ke2=Ke1' + Ke2' again Ke2=0 and 1/2 term can be dropped

m1U1^2= m1V1^2 + m2V2^2 (See ref 1)
solving these equations we get

V1=(m1-m2)U1/(m1+m2) and since U1=sqrt(2 gh)= 7.2 m/s
V1=(0.550 - 1.00) (7.2)/(550 +1.00)
V1=-2.1m/s

V2= 2m1U1/(m1+m2)

b) Again Ke(1/2)mV^2=mgh
h=(1/2)V^2 /g= 0.5 x (2.1)^2 /9.81=.222 m

c) Time of fall =time of horizontal travel. Horizontal velocity is constant
t=sqrt(2 h2 / g)
t= sqrt( 2 x 2.2 / 9.81)=0.67 s
S= V2 t= 2m1U1/(m1+m2) t=
S=[2 x 0.550 x 7.2/(0.55 +1.00) ] 0.67 =
S=3.42m

d) The horizontal velocity will be the same in magnitude as it was after m1 colided with m2. The time of flight is the sam efor both of them so

S= V1t= 2.1 x 0.67= 1.41 m

Have fun!

Answer 2

I will walk you through how to solve this problem conceptually, and I will give you the correct answers for each part. But you will have to find the correct formulas and work it out on your own to confirm the answers and to practice the work.

(a) First, convert the potential energy of m1 to its kinetic energy just before impacting m2. Then you need to use the conservation of momentum (before and after the collision must be equal) and the conservation of energy (before and after the collision must be equal) simultaneously to solve for the velocity of each mass after the collision. Note that the sign of the velocity of m1 will change.
V1 = 2.09m/s
V2 = 5.12m/s

(b) Now use the conservation of energy to convert the kinetic energy of m1 after the collision back to a potential energy of m1 to find the height back up the ramp.
h = 0.223m

(c) Now you need to use the kinematic equations to find out how far m2 flies away from the table in the time that it takes to fall to the ground. First, calculate the time it will take to fall vertically. (t = 0.67s) Now you can calculate how far it will travel horizontally at its post-collision velocity in that amount of time.
x2 = 3.43m

(d) Since the ramp is frictionless, m1 will have the same velocity after sliding back down the ramp as it did immediately after the collision. Use the post-collision velocity of m1 to calculate the distance exactly as you did for (c).
x1 = 1.40m