Home > Physics > Conservation of Momentum

A ball with a mass of #5 kg# moving at #3 m/s# hits a still ball with a mass of #6 kg#. If the first ball stops moving, how fast is the second ball moving?

Answer 1

Amelia Adams

I found #2.5m/s#

We can use Conservation of Linear Momentum #p=mv# along the horizontal #x# direction so that:

#p_("before")=p_("after")#

#(5*3)+(6*0)=(5*0)+(v*6)#

#15=6v#

#v=15/6=2.5m/s#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Charles Aeschliman

To find the final velocity of the second ball, we can use the principle of conservation of momentum:

(m_1 * v_1 + m_2 * v_2 = m_1 * v_{1f} + m_2 * v_{2f})

Where: (m_1 = 5) kg (mass of the first ball), (v_1 = 3) m/s (initial velocity of the first ball), (m_2 = 6) kg (mass of the second ball), (v_2 = 0) m/s (initial velocity of the second ball), (v_{1f} = 0) m/s (final velocity of the first ball, as it stops moving), (v_{2f}) is the final velocity of the second ball.

Solving for (v_{2f}):

(5 * 3 + 6 * 0 = 5 * 0 + 6 * v_{2f})

(15 = 6 * v_{2f})

(v_{2f} = \frac{15}{6} = 2.5) m/s

So, the second ball moves at a speed of 2.5 m/s after the collision.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.