I'm out playing near a frozen pond. On the pond is a block of ice (initially at rest) with mass 17.0 kg . I slide a different ice block of mass 2.75 kg so that it collides with the stationary block. Just before impact (long after I stopped pushing it), the moving block has speed 6.25 m/s . Ignore friction with the pond. When the blocks collide they stick together and move off as a unit.

Required:
What is the change ÎK=KfinalâKinitial in the block-snowball system's kinetic energy due to the collision?

[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\dfrac{17\times 0+2.75\times 6.25}{17+2.75}\\\Rightarrow v=0.87\ \text{m/s}[/tex]

Change in kinetic energy is given by

[tex]\Delta K=K_f-K_i\\\Rightarrow \Delta K=\dfrac{1}{2}(m_1+m_2)v^2-\dfrac{1}{2}m_2u_2^2\\\Rightarrow \Delta K=\dfrac{1}{2}\times (17+2.75)0.87^2-\dfrac{1}{2}\times 2.75\times 6.25^2\\\Rightarrow \Delta K=-46.24\ \text{J}[/tex]

THe change in kinetic energy in the system is [tex]-46.24\ \text{J}[/tex].

andromache andromache · Answer 2 · 2022-03-18T18:26:54+01:00

The change in kinetic energy in the system is -46.24 J.

Calculation of the change in the kinetic energy:

Since m1 = Mass of block at rest = 17 kg

m2 = Mass of moving block = 2.75 kg

u1 = Velocity of block at rest = 0

u2 = Velocity of moving block = 6.25 m/s

v = Velocity of the combined mass

So,

momentum in the system should be

[tex]m_1v_1 +m_2v_2 = (m_1 + m_2)v\\\\v = m_1v_1+m_2v_2\div m_1 + m_2\\\\= 17\times 0+2.75\times 6.25 \div 17 + 2.75[/tex]

= 0.87 m/s

Now the change in the kinetic energy should be

[tex]= 1\div 2 (m_1 + m_2)v^2 - 1\div 2 m_2v^2\\\\= 1\div 2\times (17 + 2.75)0.87^2 - 1\div 2\times 2.75\times 6.25^2[/tex]

= -46.24J

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