Physics education research, electronic materials, and the musings of Christopher Moore, Ph.D.
Physics education research, electronic materials, and the musings of Christopher Moore, Ph.D.
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A block is released from rest at the top of a frictionless incline plane 16 m long. It reaches the bottom 4.2 s later. A second block is projected up the plane from the bottom at the instant the first block is released in such a way that it returns to the bottom simultaneously with the first block.
a. Find the acceleration of each block on the incline.
b. What is the initial velocity of the second block?
c. How far up the incline does it travel? It can assumed that both blocks experience the same acceleration.
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S = a/2 * T^2
a= 16/(4.2^2) * 2 = 1.82 (is the accleration of the first block and the scond block too. Ans a.)
For 2nd block,
total time up the incline(TU) = total time down the incline (TD)
and TU+TD = 4.2
so TD = 4.2/2 = 2.1 = T/2
The distance it travel down within that time due to accleration 'a' is
S'= a/2 * TD^2 = a/2 * (T/2)^2 = (a/2 * T^2)/4 = S/4 = 4m ( is the distance the second block travels up the plane. Ans c.)
Now for question b. we know that for 2nd block initial velocity = final velocity. So,
V = sqrt(2*a*S) = sqrt( 2 * 1.82 * 4m) = 3.81 m/s.
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