Question

Given that the collision is elastic and glider 2 is initially at rest (v2,i =0), please use below Eqs. to explain why

1. Glider 2 will be always kicked toward the same direction as glider 1 comes in (v2,f and v1,i have the same sign)
2. Glider 1 will bounce back (v1f and v1,i have opposite sign) if it is lighter than glider 2 (m1= m2).
3. Glider 1 will keep moving forward (v1,f and v1 i have the same sign) if it is heavier than glider 2 (m1 m2)
4. Glider 1 will stop (v1f =0) if it weighs the same as glider 2 (m1= m2)


v1,f= (m1-m2)v1,i + 2m2v2,1/m1+m2
v2,f= (m2-m1)v2,i + 2m1v1,i/m1+m2

Answer 1

Answer:

Explanation:

1 )

Put v2,i =0, in second equation

v2,f= (m2-m1)v2,i + 2m1v1,i/m1+m2

v2,f = 0 + 2m1v1,i/m1+m2

v2,f =  2m1v1,i/m1+m2

In this equation coefficient of v1,i is positive so v2,f and v1,i have the same sign.

2 )

Put m1 < m2  and v2,i =0 in first equation

v1,f= (m1-m2)v1,i + 2m2v2,1/m1+m2

v1,f= (m1-m2)v1,i

As m1-m2 is negative , v1f and v1i will have opposite sign.

3 )

Put m1 > m2  and v2,i =0 in first equation

v1,f= (m1-m2)v1,i + 2m2v2,1/m1+m2

v1,f= (m1-m2)v1,i

m1 - m2 is positive so v1f and v1i will have same  sign.

4 )

Put m1 = m2 and v2,i =0 in first equation

v1,f= (m1-m2)v1,i

= 0 because m1 = m2

So glider 1 will stop because v1,f = 0 .

 

 

Answer 2

When the glider 2 is initially at rest (v2,i =0) and collision is elastic, the different equation are explained for different conditions of glider 1.

What is conservation of momentum?

Momentum of a object is the force of speed of it in motion. Momentum of a moving body is the product of mass times velocity.

When the two objects collides, then the initial collision of the two body is equal to the final collision of two bodies by the law of conservation of momentum.

The given equations are,

$v_{1f}= \dfrac{(m_1-m_2)v_{1i} + 2m_2v_{2i}}{m1+m2}$

$v_{2f}= \dfrac{(m_2-m_1)v_{2i} + 2m_1v_{1i}}{m1+m2}$

Given that the collision is elastic and glider 2 is initially at rest (v2,i =0),

• 1. Glider 2 will always be kicked toward the same direction as glider 1 comes in (v2,f and v1,i have the same sign)

Put $v_{2i}$ equal to zero, in the given equation as,

$v_{2f}= \dfrac{(m_2-m_1)(0) + 2m_1v_{1i}}{m1+m2}\\v_{2f}= \dfrac{ 2m_1v_{1i}}{m1+m2}$

v2,f and v1,i have the same sign in the above equation.

• 2. Glider 1 will bounce back (v1f and v1,i have opposite sign) if it is lighter than glider 2 (m1<m2).

Put $v_{1i}$ equal to zero, in the given equation as,

$v_{1f}= \dfrac{(m_2-m_1)(0) + 2m_2v_{2i}}{m1+m2}\\v_{1f}= \dfrac{ 2m_2v_{2i}}{m1+m2}$

Now, if we consider (m1<m2), for the above equation, the result will be negative. Thus, v1f and v1,i have opposite sign.

• 3. Glider 1 will keep moving forward (v1,f and v1 i have the same sign) if it is heavier than glider 2 (m1 >m2)

Put $v_{2i}$ equal to zero, in the first equation as,

$v_{1f}= \dfrac{(m_2-m_1)(v_{1i}) + 2m_2(0)}{m1+m2}\\v_{1f}= \dfrac{(m_2-m_1)(v_{1i})}{m1+m2}$

Now, if we consider (m1>m2), for the above equation, the result will be positive. v1,f and v1 i have the same sign.

• 4. Glider 1 will stop (v1f =0) if it weighs the same as glider 2 (m1= m2)

Put $v_{2i}$ equal to zero, in the first equation as,

$v_{1f}= \dfrac{(m_2-m_1)(v_{1i}) + 2m_2(0)}{m1+m2}\\v_{1f}= \dfrac{(m_2-m_1)(v_{1i})}{m1+m2}$

Now, if we consider (m1=m2), for the above equation, the result will be zero. Thus the glider 1 will stop (v1f =0).

Hence, when the glider 2 is initially at rest (v2,i =0) and collision is elastic, the different equation are explained for different conditions of glider 1.

Learn more about the conservation of momentum here;

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