Ryan H. answered 02/22/26
PhD in Physics with 3 years teaching and tutoring experience
This is a multistep question that builds on multiple prior subjects.
You are initially given a scenario of a collision resulting in the two blocks falling off a table at different speeds, and are given how far the blocks fall from the table. To get to the answer, the speed of the large block before the collision, you need to work backwards. Start with the top of the table, and remember that objects like this will have the same acceleration downward as an object dropped from this height. Use this to find out how long they were in the air before hitting the ground. We'll call that T for time.
Y = 0.5aT2 -> T = root(2Y/a)
Y = 1.10 m, a = 9.8 m/s2
Then, using the distance from the edge of the table and your kinematics equation at net 0 acceleration, you can figure out the speeds of the big block, call it M1 and v1, and the small block, M2 and v2, based on their distances d1 and d2 and the time calculated before, T.
M1: d1 = v1*T -> v1 = d1/T = d1/root(2Y/a)
M2: d2 = v2*T -> v2 = d2/T = d2/root(2Y/a)
Now that you have v1 and v2, you know that momentum is conserved before and after collisions. The initial velocity of the small block is 0, and the answer you are after is the initial velocity of M1, let's call it v0.
M1 * v0 = M1v1 + M2v2 -> v0 = (M1v1 + M2v2) / M1
Fill in the values for M1 and M2 as given in the problem, and v1 and v2 as derived above. When I plug in the numbers, I get a pre-collision speed of 5.8 m/s.