Question

A Big fish eats a little fish. The large 45.0 kg fish is moving at a velocity of
10.0 m/s when it swallows a 5.0 kg fish at rest. How fast will the two fish be
going after the collision?​ ​

Answer 1

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ Given :- }}}}$

• A Big fish eats a little fish. The large 45.0 kg fish is moving at a velocity of 10.0 m/s when it swallows a 5.0 kg fish at rest.

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ To \: Find :- }}}}$

• How fast will the two fish be going after the collision .

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ Solution :- }}}}$

Given that a Big fish eats a little fish. The large 45.0 kg fish is moving at a velocity of 10.0 m/s when it swallows a 5.0 kg fish at rest.We need to find how fast will the two fish be going after the ccollision . We can use the law of conservation of momentum according to which in a system the Total momentum remains conserved .

★Using the Law of Conservation of Momentum:-

$\begin{gathered}\sf\dashrightarrow \pink{ m_1u_1+m_2u_2= m_1 v_1+m_2v_2 }\\\\\sf\dashrightarrow m_1u_1+m_2u_2 = v ( m_1+m_2) \\\\\sf\dashrightarrow (45 kg)(10m/s)+(5 kg)(0m/s)= v(45+5) \\\\\sf\dashrightarrow 450 kg-m/s + 0 = (50kg)v \\\\\sf\dashrightarrow v(50kg) = 450 kg-m/s \\\\\sf\dashrightarrow v =\dfrac{450 kg-m/s}{50kg} \\\\\sf\dashrightarrow \underset{\blue{\sf Required\ Velocity }}{\underbrace{\boxed{\pink{\frak{ Velocity_{(combined)} = 9m/s}}}}}\end{gathered}$

Answer 2

Answer:

★ Given :

A big fish eats a little fish.

The large 45.0 kg fish is moving at a velocity of 10.0 m/s when it swallows a 5.0 kg fish at rest.

★ To Find :

How fast will the two fish be going after the collision .

★ Solution :

Given that a big fish eats a little fish.

The large 45.0 kg fish is moving at a velocity of 10.0 m/s when it swallows a 5.0 kg fish at rest.

We need to find how fast will the two fish be going after the collision .

We can use the law of conservation of momentum according to which in a system the total momentum remains conserved .

▶ Using the Law of Conservation of Momentum :

m1u1 + m2u2 = m1v1 + m2v2

• m1u1 + m2u2 = v ( m1+ m2 )

• (45kg) (10m/s) + (5kg) (0m/s) = v ( 45kg + 5kg )

• 450kgm/s + 0 = v (50kg)

• 450kgm/s = v (50kg)

• v = 450kgm/s / 50kg

• v = 9m/s

Hence : Velocity = 9 m/s