Question

pleaseeee help. only three questions will award medals and fan !!
1.) Ball A with a mass of 20 kg is moving to the right at 20m/s. At what velocity should Ball B, with a mass of 40 kg move so that they both come to a stand still?
2.) Ball B with a mass of 30 kg is moving to the left at 10 m/s. with what velocity should Ball A with a mass of 10 kg move to the right and collide with Ball B so that Ball A rebounds with a velocity of 30 m/s, and Ball B with a velocity of 10 m/s after the collision?
3.) Ball A with a mass of 10 kg is moving to the right at 20 m/s. Ball B is moving to the left

Answer 1

Ball B is moving to the left at 20 m/s. Upon collision, Ball B comes to a standstill and Ball A moves to the left at twice its original speed. What is the mass of Ball B?

Answer 2

@mathslover @ganeshie8 @Destinymasha @Jemurray3 @esshotwired can anyone of you help?? or atleast explain what it is I need to do to get the answer.

Answer 3

@secret66

Answer 4

@MrNood ???

Answer 5

This is a conservation of momentum problem. The momentum of each ball is its mass times it's velocity - \(p= mv\). The momentum of the total system is the sum of the momenta of each ball - \(p_{system} = p_1 + p_2 = m_1 v_1 + m_2 v_2 \).

In each of those, you are given the velocity of each ball after the collision - which means that you can calculate \( p_{system}\) after the collision. The conservation of momentum states that the momentum of the system beforehand is equal to the momentum afterward. Therefore, you are supposed to find the unknown velocity of the second ball by setting the initial momentum equal to the final momentum and solving.

So for the second one:

\[ m_A = 10 kg, \space m_B = 30 kg\]
after the collision,
\[ v_A = 30 m/s, \space v_B = 10 m/s\]
which means that after the collision,
\[ p_{after} = m_A v_A + m_B v_B = (10 kg)(30 m/s) + (30kg)(10 m/s) = 600 kg m/s\]

Before the collision,
\[ v_A = x, \space v_B = -30 m/s\]
where we have not yet found x, and the minus sign is because it's moving to the left. The initial momentum is
\[ p_{before} = m_A v_A + m_B v_B = (10kg)x - (30kg)(30m/s) = 10x - 900 kg m/s\]

Momentum conservation says these two things are equal:

\[ 600 kgm/s = 10x - 900 kgm/s\]
so
\[10x = 1500 kg m/s\]
\[ x = 150 m/s\]

Answer 6

Actually, that's not the second one - I mixed up \( v_A \) and \(v_B\). But that's an example of how to use the conservation of momentum to solve a problem like this. Use that procedure and use the right numbers to get your answer.