Q12Q Consider a head-on collision bet... [FREE SOLUTION]

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Chapter 10: Q12Q (page 410)

Consider a head-on collision between two objects. Object 1, which has mass $m_{1}$ , is initially in motion, and collides head-on with object 2, which has mass $m_{2}$ and is initially at rest. Which of the following statements about the collision are true?
(1) ${\overset{鈫�}{p}}_{1,initial} = {\overset{鈫�}{p}}_{1, final} + {\overset{鈫�}{p}}_{2, final}$ .
(2) $| {\overset{鈫�}{p}}_{1, final} | < | {\overset{鈫�}{p}}_{1, initial} |$ .
(3) If $m_{2} 鈮� m_{1}$ , then $| 螖 {\overset{鈫�}{p}}_{1} | > | 螖 {\overset{鈫�}{p}}_{2} |$ .
(4) If $m_{1} 鈮� m_{2}$ , then the final speed of object 2 is less than the initial speed of object 1.
(5) If $m_{2} 鈮� m_{1}$ , then the final speed of object 1 is greater than the final speed of object 2.

Short Answer

Expert verified

The correct statements are (1), (2) and (5).

Step by step solution

Given data

A head on collision of object 1 and object 2 having mass $m_{1}$ and $m_{2}$ which were in rest.

The concept of momentum principle

A net force affects the momentum of an object whose momentum is the product of mass and velocity, according to the momentum principle.

$F_{net} = \frac{螖 p}{螖 t}$

Step 3:The correct statements and the reasons

The statement (1) is correct as:

${\overset{鈫�}{p}}_{1, initial} = {\overset{鈫�}{p}}_{1, final} + {\overset{鈫�}{p}}_{2, final}$

Collision of will follow the momentum principle. So, the momentum before collision is equal to the momentum after collision, and ${\overset{鈫�}{p}}_{2, initial} = 0$ .

The statement (2) is correct as:

$|{\overset{鈫�}{p}}_{1, final}| < |{\overset{鈫�}{p}}_{1, initial}|$

From the momentum principle, the amount of ${\overset{鈫�}{p}}_{1, initial}$ is equal to ${\overset{鈫�}{p}}_{1, final} + {\overset{鈫�}{p}}_{2, final}$

The statement (2) is correct.

(3) The statement is, $m_{2} > m_{1}$ , then $|螖 P_{1}| > |螖 P_{2}|$ .

Break it down into terms so we can see if the equation is true.

$|螖 P_{1}| = |螖 P_{2}| m_{1} (v_{1} - u_{1}) = m_{2} (v_{2} - u_{2}) m_{1} (v_{1} - u_{1}) = m_{2} (v_{2})$

If $m_{2} > m_{1}$ , then $|螖 P_{1}| < 鈭� 螖 P_{2}$ , which means the statement is not true.

(4) Let us check if the final speed of object 2 is less than the initial speed of the object

$|螖 P_{1}| = |螖 P_{2}| m_{1} (v_{1} - u_{1}) = m_{2} (v_{2} - u_{2}) m_{1} (v_{1} - u_{1}) = m_{2} (v_{2}) v_{2} = \frac{m_{1} (v_{1} - u_{1})}{m_{2}}$

Here, if $m_{1} > m_{2}$ , the final speed of the second object is greater than the first object's initial speed. This will make the statement, not true.

(5) Let us check if the final speed of object 1 is greater than the final speed of the object 2 .

$|螖 P_{1}| = |螖 P_{2}| m_{1} (v_{1} - u_{1}) = m_{2} (v_{2} - u_{2}) m_{1} (v_{1} - u_{1}) = m_{2} (v_{2}) v_{2} = \frac{m_{1} (v_{1} - u_{1})}{m_{2}}$