PHysics  ONLY question 2 b) so what I would like you to do is read the questions and  read the feedback and fix my answers thats all    here is the question :    Here is the feedback :  You still did not take into consideration that the combined masses are moving when the spring is compressed. Find speed of combined mass at maximum compression using Momentum equation PTI = PTF and then use Energy Equations to solve for the spring constant.  Ek1 + Ek2 = EkT + EE  Where EE is the elastic energy stored in the spring. Then use EE = 0.5kx2 to solve for k

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Author:Raymond A. Serway, John W. Jewett
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Chapter8: Momentum And Collisions
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PHysics 

ONLY question 2 b)

so what I would like you to do is read the questions and  read the feedback and fix my answers thats all 

 

here is the question : 

 

Here is the feedback : 

You still did not take into consideration that the combined masses are moving when the spring is compressed. Find speed of combined mass at maximum compression using Momentum equation PTI = PTF and then use Energy Equations to solve for the spring constant.  Ek1 + Ek2 = EkT + EE  Where EE is the elastic energy stored in the spring. Then use EE = 0.5kx2 to solve for k

here is my answer : fix it correctly and give me the final answer : 

1. A ball with a mass of 10.0 g hits a 5.0 kg block of wood tied to a
string, as shown. You may assume that there is no friction from the
string. The ball hits the block in a completely inelastic collision, causing
the block to rise to a maximum height of 8.0 cm. Find the initial
horizontal speed of the ball. You may assume that the string is very
long.
|||
Ah = 8.0 cm
Press here for long description
2.A 1.2 kg cart moving at 6.0 m/s [E] collides with a stationary 1.8 kg
cart. The head-on collision is completely elastic and is cushioned by a
spring.
a. Find the velocity of each cart after the collision.
b. The maximum compression of the spring during the collision is 2.0
cm. Find the spring constant.
Transcribed Image Text:1. A ball with a mass of 10.0 g hits a 5.0 kg block of wood tied to a string, as shown. You may assume that there is no friction from the string. The ball hits the block in a completely inelastic collision, causing the block to rise to a maximum height of 8.0 cm. Find the initial horizontal speed of the ball. You may assume that the string is very long. ||| Ah = 8.0 cm Press here for long description 2.A 1.2 kg cart moving at 6.0 m/s [E] collides with a stationary 1.8 kg cart. The head-on collision is completely elastic and is cushioned by a spring. a. Find the velocity of each cart after the collision. b. The maximum compression of the spring during the collision is 2.0 cm. Find the spring constant.
Mass of cart 1 (m₁) = 1.2 kg
Initial velocity of cart 1 (v₁₁) = 6.0 m/s
Mass of cart 2 (m₂) = 1.8 kg
Initial velocity of cart 2 (V₂i) = 0 m/s
Maximum compression of the spring (x) = 2 cm = 0.02 m
The maximum compression of the spring occurs when all the kinetic energy is converted into
potential energy in the spring. The potential energy stored in a spring is given by
P.E. =-=kx²
where k is the spring constant and x is the displacement (compression in this case).
The potential energy stored in the compressed spring is equal to the initial kinetic energy of the
system,
The potential energy of the spring = Initial kinetic energy
1/1/2xKxx²=
= 1/2 xm₁ × (v₁,;)² + 1/{ xm₂×(√₂₁)²
X
11/12×
-×k×(0.02 m)² = —×(1.2 kg)× (6.0 m / s)² + 1 ×(1.8 kg)× (0 m/
2
(0.0002 m²) xk = 21.6 J+ 0 J
21.6 J
0.0002 m²
k= 108000 N/m
(b) The spring constant is 108000 N/m.
K=
Transcribed Image Text:Mass of cart 1 (m₁) = 1.2 kg Initial velocity of cart 1 (v₁₁) = 6.0 m/s Mass of cart 2 (m₂) = 1.8 kg Initial velocity of cart 2 (V₂i) = 0 m/s Maximum compression of the spring (x) = 2 cm = 0.02 m The maximum compression of the spring occurs when all the kinetic energy is converted into potential energy in the spring. The potential energy stored in a spring is given by P.E. =-=kx² where k is the spring constant and x is the displacement (compression in this case). The potential energy stored in the compressed spring is equal to the initial kinetic energy of the system, The potential energy of the spring = Initial kinetic energy 1/1/2xKxx²= = 1/2 xm₁ × (v₁,;)² + 1/{ xm₂×(√₂₁)² X 11/12× -×k×(0.02 m)² = —×(1.2 kg)× (6.0 m / s)² + 1 ×(1.8 kg)× (0 m/ 2 (0.0002 m²) xk = 21.6 J+ 0 J 21.6 J 0.0002 m² k= 108000 N/m (b) The spring constant is 108000 N/m. K=
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