A ball with a mass of #2 kg# moving at #16 m/s# hits a still ball with a mass of #21 kg#. If the first ball stops moving, how fast is the second ball moving? How much kinetic energy was lost as heat in the collision?

Answer 1

In the first scenario, momentum conservation law will be applied.

from the law, #m_1u_1+m_2u_2=m_1v_1+m_2v_2#
here, #m_1=2kg# #m_2=21kg# #u_1=16ms^(-1)# #u_2=0ms^(-1)# #v_1=0ms^(-1)# we have to find #v_2# if we put all the values in the equation, we will find,
#2*16+21*0=2*0+21*v_2# #or, 32=21v_2# #or, v_2=32/21# #or, v_2=1.524ms^(-1)#
for the energy loss, #E_(i)=1/2*2*16^2+1/2*21*0^2# #=256j#
#E_(f)=1/2*2*0^2+1/2*21*1.524^2# #=24.387j#
so, #Delta E=E_(i)-E_f# #=256j-24.387j# #=231.613j#
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Answer 2

To find the velocity of the second ball after the collision:

v2_final = (m1 * v1_initial) / m2 v2_final = (2 kg * 16 m/s) / 21 kg v2_final ≈ 1.5238 m/s

To find the kinetic energy lost as heat in the collision:

Initial kinetic energy: KE_initial = 0.5 * m1 * v1_initial^2 + 0.5 * m2 * v2_initial^2 KE_initial = 0.5 * 2 kg * (16 m/s)^2 + 0.5 * 21 kg * (0 m/s)^2 KE_initial = 256 J + 0 J KE_initial = 256 J

Final kinetic energy: KE_final = 0.5 * m1 * v1_final^2 + 0.5 * m2 * v2_final^2 KE_final = 0.5 * 2 kg * (0 m/s)^2 + 0.5 * 21 kg * (1.5238 m/s)^2 KE_final = 0 J + 23.0664 J KE_final ≈ 23.0664 J

Kinetic energy lost = 256 J - 23.0664 J Kinetic energy lost ≈ 232.9336 J

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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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