A m1 = 0.500 kg block is released from rest at the top of a frictionless track h

Question

A m1 = 0.500 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 the figure below.

(a) Determine the velocities of the two objects just after the collision. velocity of m1 1 m/s
velocity of m2 2 m/s
(b) How high up the track does the 0.500 kg object travel back after the collision?
3 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?
4 m
(d) How far away from the bottom of the table does the 0.500 kg object eventually land?
5 m

Explanation / Answer

2: Conservation of linear momentum due to being an isolated system. 1: Conservation of kinetic energy bc it's an elastic collision: 1: .5mv^2 of m1 right before it hits block m2 is given by mgh=KE (so set mgh=.5mv^2) Bc KE is conserved: .5mv^2 of m1 + .5mv^2 of m2 after the collision equal the sum of the KE's before the collision which was equal to mgh as per above (the KE of m2 was initially 0) So: v^2 f1 (the final velocity of block 1, squared) = v^2 01 (the initial v of block 1, squared) - m2/m1*v^2f2 Then use conservation of linear momentum (m1v1 + m2v2 (final) = m1v1 (initial) to find vf2 = m1/m2(v01-vf1) Put that into the equation above to yield: v^2 f1=v^2 01 - m2/m1*vf2=v^2 01-m2/m1[m1/m2(v01-vf1)]^2 Solve for vf1 and you see that: vf1= [(m1-m2)/(m1+m2)]*v01 Substitute this into the equation for vf2 listed above with the cons of linear momentum and you can solve for vf2. Once you have these, you can solve for the KE of the .5 kg block after the collision, set that equal to mgh again and solve for h for question b. You can use simple linear dynamics to solve c and d given the velocities will remain the same as they travel on the frictionless table. KE for d will be the same as the KE just after the collision.

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