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deff fn [24]
3 years ago
13

A 15 kg block is sliding along a frictionless surface and strikes a 10 kg ball at rest. What is the collision of the blocks afte

r the collision if it is an inelastic collision?
Physics
1 answer:
vfiekz [6]3 years ago
7 0

Answer:

   v = 0.6 v₁

Explanation:

This is an exercise in collisions, let's start by defining a system formed by the two bodies, so that the forces during the collisions have been internal and the momentum is preserved.

Instant starts. Before the crash

       p₀ = M v₁ + m 0

Final moment. After the crash

       p_{f} = (M + m) v

how momentum is conserved

       p₀ = p_{f}

       M v₁ = (M + m) v

       

       v = \frac{M}{M+m} v_{1}

let's calculate

       v = \frac{15}{15+10} v_{1}

        v = 0.6 v₁

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kenny6666 [7]

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A force of 44 N will stretch a rubber band 88 cm ​(0.080.08 ​m). Assuming that​ Hooke's law​ applies, how far will aa 11​-N forc
Setler79 [48]

Answer:

<em>The rubber band will be stretched 0.02 m.</em>

<em>The work done in stretching is 0.11 J.</em>

Explanation:

Force 1 = 44 N

extension of rubber band = 0.080 m

Force 2 = 11 N

extension = ?

According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.

F = ke

where k = constant of elasticity

e = extension of the material

F = force applied.

For the first case,

44 = 0.080K

K = 44/0.080 = 550 N/m

For the second situation involving the same rubber band

Force = 11 N

e = 550 N/m

11 = 550e

extension e = 11/550 = <em>0.02 m</em>

<em>The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch</em>. This is in line with energy conservation.

potential energy stored = \frac{1}{2}ke^{2}

==> \frac{1}{2}* 550* 0.02^{2} = <em>0.11 J</em>

3 0
3 years ago
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