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(a) Time of collision: The time it takes for the blocks to collide is the same as the time it takes for Block 1 to fall 1 meter. We can use the equation of motion [tex]$d = ut + \frac{1}{2}gt^2$$[/tex], where d is the distance, u is the initial velocity, g is the acceleration due to gravity, and t is the time.
Substituting the given values (Block 1 falls from rest, so u = 0, and d = 1 m, we get:
[tex]$1 = \frac{1}{2}(9.8)t^2$$[/tex]
Solving for t, we get [tex]$$t \approx 0.45 s$$[/tex].
(b) Height of collision: The height at which the collision occurs is the height reached by Block 2 in the time it takes for the blocks to collide. We can use the equation of motion [tex]$h = ut - \frac{1}{2}gt^2$$[/tex]
Substituting the given values (Block 2 is launched with an initial velocity of 10 m/s, and t = 0.45 s), we get:
[tex]$h = 10(0.45) - \frac{1}{2}(9.8)(0.45)^2 \approx 4.0 m$$[/tex]
(c) Speeds just before collision: Just before the collision, Block 1 will have fallen for 0.45 s and Block 2 will have been rising for the same amount of time. We can use the equation v = u + gt to find their speeds,
For Block 1 (which falls from rest), we get:
[tex]$v_1 = 0 + 9.8(0.45) \approx 4.4 m/s$$[/tex]
For Block 2 (which is launched with an initial velocity of 10 m/s), we get:
[tex]$v_2 = 10 - 9.8(0.45) \approx 5.6 m/s$$[/tex]
(d) Highest point after collision: After the collision, the blocks move together with a combined mass of 4 kg and a combined velocity equal to their average velocity just before the collision, which is [tex]$\frac{v_1 + v_2}{2} \approx 5.0 m/s$$[/tex].
The highest point they reach can be found using the equation [tex]$h = \frac{v^2}{2g}$$[/tex],
Substituting the given values, we get:
[tex]$h = \frac{(5.0)^2}{2(9.8)} \approx 1.3 m$$[/tex]
(e) Minimum launch speed for 1 m height after collision: If we want the blocks to reach a height of 1 m after the collision, we can use the equation [tex]$h = \frac{v^2}{2g}$$[/tex] to find the required initial velocity.
Substituting the given values (and solving for v), we get:
[tex]$v = \sqrt{2gh} = \sqrt{2(9.8)(1)} \approx 4.4 m/s$$[/tex]
This is the average velocity of the blocks just before the collision, so the minimum launch speed for Block 2 is twice this value (since Block 1 starts from rest), or approximately 8.8 m/s.
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