Question

A 3 kg object moving at 3 m/s hits a 2 kg object at rest. If they stick together, what is the velocity of the combined mass?

Answer 1

The velocity of the combined mass is 1.8 m/s.

Step-by-step explanation:

In this collision problem, we can use the principle of conservation of momentum to find the velocity of the combined mass.

The principle of conservation of momentum states that the total momentum of a system remains constant before and after a collision, as long as no external forces are acting on the system.

Before the collision, the momentum of object A is calculated as:

1. Momentum of object A = mass of object A × velocity of object A
2. Momentum of object A = 3 kg × 3 m/s
3. Momentum of object A = 9 kg m/s

Since object B is initially at rest, its momentum is:

1. Momentum of object B = mass of object B × velocity of object B
2. Momentum of object B = 2 kg × 0 m/s
3. Momentum of object B = 0 kg m/s

After the collision, the two objects stick together, so they have the same velocity. Let's assume this velocity is v. Using the principle of conservation of momentum:

1. Total momentum before the collision = Total momentum after the collision
2. Momentum of object A + Momentum of object B = (Total mass of object A and B) × v
3. 9 kg m/s + 0 kg m/s = (3 kg + 2 kg) × v
4. 9 kg m/s = 5 kg × v
5. v = 9 kg m/s / 5 kg
6. v = 1.8 m/s

Therefore, the velocity of the combined mass after the collision is 1.8 m/s.

Answer 2

The velocity of the combined mass is 1.8 m/s.

Step-by-step explanation:

We have,

Mass of object 1,
`m_1=3\ kg`

Initial speed of object 1,
`u_1=3\ m/s`

Mass of object 2,
`m_2=2\ kg`

Initial speed of object 2 is 0. It was at rest initially.

It is required to find the velocity of the combined mass after the collision, if they stick together. The momentum of the system remains conserved. Using conservation of momentum as :

`m_1u_1+m_2u_2=(m_1+m_2)V`

V is common velocity

`V=(m_1u_1)/(m_1+m_2)\\\\V=(3* 3)/(3+2)\\\\V=1.8\ m/s`

So, the velocity of the combined mass is 1.8 m/s.