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4. Since the system is motionless before the string is burned, by conservation of momentum the total momentum of the blocks must remain zero after the string is burned, i.e. we have
m1 v1 + m2 v2 = 0.
Thus we have
v1 = -m2 v2/m1 = -(7.6)(4.6)/(0.28) = -124.9 m/s,
the minus sign meaning that v1 points to the left as shown. Now by conservation of energy, the potential energy E stored in the spring before the string is burned must be equal to the total kinetic energy of the blocks after the string is burned, i.e.
E = 1/2 (m1 v1^2 + m2 v2^2)
= 1/2[(0.28)(124.9)^2 + (7.6)(4.6)^2]
= 2260 J
All units are MKS, so there is no need to write them down in the calculation.
2. Note: The answer to this problem is not well-defined because it depends on where on the incline the portion with friction is located. We will assume that it is located at the top of the incline. Otherwise, the answer will be greater since the block will enter the portion with friction later and thus have a greater speed when decelerated by friction, and thus this constant deceleration will occur over a shorter period of time, slowing down the block less.
Let v1 be the speed of the block immediately after it leaves the spring and let v2 be its speed immediately after it leaves the portion of the incline with friction. By conservation of energy, we have
1/2 m v1^2 = 1/2 k x^2,
where x = 0.35 cm is the initial compression of the spring. Thus we have
v1 = x sqrt(k/m) = (0.35)sqrt(310/2) = 4.36 m/s.
Now as ...
The solution solves various problems in Newtonian mechanics.