From the calculations, the final momentum of B is 8.16 m/s
<h3>What is conservation of linear momentum?</h3>
According to the principle of the conservation of linear momentum, the momentum before collision is equal to the total momentum after collision.
This implies that;
MaUa + MbUb = MaVa + MaVa
Substituting values;
(0.08 kg * 0.5 m/s) + (0.05 kg * 0 m/s) = (0.08 kg * −0.1 m/s) + (0.05 kg * v)
0.4 = -0.008 + 0.05v
v = 8.16 m/s
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Answer:
B v1= 12.5m/s
Explanation:
As the acceleration is constant and time taken is also same and distance covered is halfed so the speed will also be halfed.
Work done on the crate is 1411.2 J
Explanation:
Work done is defined as the product of force and the distance moved by the object. The unit of work done is in joules and denoted by the symbol J.
Work done = F * d
where F represents the force and d represents the distance moved by the object.
mass = 72 kg , distance moved by the object is given by 2.0 m
Force F = mass * gravity = 72 * 9.8
= 705.6 N =706 N.
Work done = 706 * 2.0 = 1412 J.
Based on the calculations, the acceleration of this block as it slides down from the friction less ramp is equal to 0.85 m/s².
<h3>How to determine the acceleration of the block?</h3>
In order to determine the acceleration of this block, we would apply Newton's Second Law of Motion while taking moment about the tensional force.
The vertical component of the tensional force that is acting on the block as it slides down from this friction less ramp is given by:
T = Ma - Mgsinθ
0 = Ma - Mgsinθ
Ma = Mgsinθ
a = gsinθ
Substituting the given parameters into the formula, we have;
a = 9.8 × sin25
a = 9.8 × 0.4226
Acceleration, a = 0.85 m/s².
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