A ball with a mass of #8 kg# moving at #4 m/s# hits a still ball with a mass of #3 kg#. If the first ball stops moving, how fast is the second ball moving?

Answer 1

#v_1^'=20/11 m/s#

#v_2^'=64/11 m/s#

#"Before collision:"# #...........................................................# #vec P_1=m_1*vec v_1" (momentum for the first object)"#
#vec P_1=8*4=32 kg*m/s#
#vec P_2=m_2*vec v_2"(momentum for the second object)"#
#vec P_2=3*0=0#
#Sigma vec P_b=vec P_1 +vec P_2# #Sigma vec P_b=32+0=32" " kg*m/s#
#"After collision :"# #".............................................................."#
#vec P_1^'=m_1*vec v_1^'"# #vec P_2^'=m_2*vec v_2^'#
#Sigma vec P_a=vec P_1^'+vec P_2^'# #Sigma vec P_a=8*v_1^'+3*v_2^'#
#"Conservation of momentum:"# #...............................................................# #Sigma vec P_b=Sigma vec P_a#
#32=8v_1^'+3v_2^'" "(1)"#
#v_1+v_1^'=v_2+v_2^'#
#4+v_1^'=0+v_2^' " "(2)#
#32=8v_1^'+3(4+v_1^')#
#32=8v_1^'+12+3v_1^'#
#32-12=11v_1^'" ; "20=11v_1^'#
#v_1^'=20/11 m/s#
#"using equation (2):"#
#4+20/11=v_2^'#
#v_2^'=64/11 m/s#
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Answer 2

To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum of a system remains constant if no external forces act on it.

The momentum (p) of an object is given by the product of its mass (m) and velocity (v): p = m * v.

Initially, the first ball has a momentum of (8 kg) * (4 m/s) = 32 kg·m/s.

When the first ball hits the second ball, they stick together, and the system is now a combined mass of 8 kg (from the first ball) + 3 kg (from the second ball) = 11 kg.

Since momentum is conserved, the total momentum after the collision is equal to the total momentum before the collision.

Therefore, the momentum of the combined system after the collision is also 32 kg·m/s.

Now, we can find the velocity of the combined system (both balls stuck together) after the collision:

Total momentum after collision = Total mass * Combined velocity

32 kg·m/s = 11 kg * Combined velocity

Combined velocity = 32 kg·m/s / 11 kg ≈ 2.91 m/s

Thus, the second ball, which is part of the combined system after the collision, is moving at approximately 2.91 m/s.

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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.

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.

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.

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.

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