Question

A 24.6 g marble sliding to the right at 62.0 cm/s overtakes and collides elastically with a 12.3 g marble moving in the same direction at 15.5 cm/s. After the collision, the 12.3 g marble moves to the right at 77.5 cm/s. Find the velocity of the 24.6 g marble after the collision. cm/s

Answer 1

Answer:

Velocity of the 24.6 g marble is 31 cm/s.

Step-by-step explanation:

Given:

Marble 1:

Mass of the marble $(m_1)$ = 12.3 gm

Velocity of $m_1$ before collision $v_1_i$ = 15.5 cm/s

Velocity of $m_1$ after collision $v_1_f$ = 77.5 cm/s

Marble 2:

Mass of the marble $(m_2)$ = 24.6 gm

Velocity of $m_2$ before collision $v_2_i$ = 62 cm/s

Velocity of $m_2$ after collision $v_2_f$ = ?

From the law of conservation of momentum, we know that momentum before equals the momentum after :

So,

⇒ $P(i)=P(f)$

⇒ $(m_1\times v_1_i )+(m_2\times v_2_i) =(m_1\times v_1_f)+(m_2\times v_2_f)$

⇒ Plugging the values.

⇒ $(12.3\times 15.5) +(24.6\times 62)=(12.3\times 77.5)+(24.6\times v_2_f)$

⇒ $190.65+1525.2=953.25+24.6\times v_2_f$

⇒ $1715.85=953.25+24.6\times v_2_f$

⇒ $1715.85-953.25=24.6\times v_2_f$ ...subtracting both sides 953.25

⇒ $762.6=24.6\times v_2_f$

⇒ $\frac{762.6}{24.6}\ =v_2_f$ ...dividing both sides with 24.6

⇒ $31=v_2_f$

⇒ $v_2_f =31$ cm/s

The velocity of the 24.6 g marble after the collision is 31 cm/s and it will move opposite to of the 12.3 g marbles that is towards left.