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mr_godi [17]
4 years ago
8

Which is true about inelastic collisions: a. An inelastic collision does not obey conservation of energy. b. An inelastic collis

ion conserves kinetic energy. c. Objects will stick together upon collision. d. Momentum is not conserved in inelastic collisions..
Physics
1 answer:
Phoenix [80]4 years ago
5 0

Answer:

Option c is correct

Explanation:

There are two types of collisions-elastic collision and inelastic collision.

In elastic collision, both kinetic energy and total momentum are conserved. On the other hand, in inelastic collision, total momentum is conserved but kinetic energy is not conserved. Thus, option b and d are incorrect.

Total energy is always conserved in both types. Thus, option a is incorrect.

In a perfectly inelastic collision, objects stick together. This happens because maximum kinetic energy is dissipated and used in bonding of the two objects. Thus, correct option is c.

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Let <em>B</em> = <em>a</em> <em>i</em> + <em>b</em> <em>j</em> + <em>c</em> <em>k</em>. Then the cross product of <em>A</em> = <em>i</em> + 2<em>j</em> - <em>k</em> with <em>B</em> is

<em>A</em> × <em>B</em> = ( <em>i</em> + 2<em>j</em> - <em>k </em>) × ( <em>a</em> <em>i</em> + <em>b</em> <em>j</em> + <em>c</em> <em>k</em> )

<em>A</em> × <em>B</em>  = <em>a</em> ( <em>i</em> × <em>i</em> ) + 2<em>a</em> ( <em>j</em> × <em>i</em> ) - <em>a</em> ( <em>k</em> × <em>i </em>)

… … … + <em>b</em> ( <em>i</em> × <em>j</em> ) + 2<em>b</em> ( <em>j </em>× <em>j</em> ) - <em>b</em> ( <em>k</em> × <em>j</em> )

… … … + <em>c</em> ( <em>i</em> × <em>k</em> ) + 2<em>c</em> ( <em>j</em> × <em>k</em> ) - <em>c</em> ( <em>k</em> × <em>k</em> )

<em>A</em> × <em>B</em> = 0 - 2<em>a</em> <em>k </em>- <em>a</em> <em>j</em>

… … … + <em>b</em> <em>k</em> + 0 + <em>b</em> <em>i</em>

… … … - <em>c</em> <em>j</em> + 2<em>c</em> <em>i</em> - 0

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So we have

3 <em>i</em> - <em>j</em> + 5 <em>k </em>= (<em>b</em> + 2<em>c</em>) <em>i</em> + (<em>c</em> - <em>a</em>) <em>j</em> + (<em>b</em> - 2<em>a</em>) <em>k</em>

which gives us the system of equations,

{ <em>b</em> + 2<em>c</em> = 3

{ -<em>a</em> - <em>c</em> = -1

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Solve for <em>a</em>, <em>b</em>, and <em>c</em>.

• Eliminate <em>c</em> from the first two equations:

(<em>b</em> + 2<em>c</em>) + 2 (-<em>a</em> - <em>c</em>) = 3 + 2 (-1)

-2<em>a</em> + <em>b</em> = 1

But -2<em>a</em> + <em>b</em> = 5, and 5 ≠ 1, so there is no such vector <em>B</em> that satisfies the cross product!

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