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A 2 kg piece of ice moving in the +x direction at 4.0
m/sstrikes a 3 kg piece of ice that is initially at rest. After
thecollision, the 2 kg piece moves at an angle of 40degrees above
the+x axis at a speed of 1.9 m/s. Find magnitude and direction of
thevelocity of the 4 kg piece.

Answer :

Answer:

Explanation:

[tex]m_1=2 kg[/tex]

velocity of first piece [tex]u_1=4 m/s[/tex]

mass of second piece [tex]m_2=3 kg[/tex]

Velocity of second piece [tex]u_2=0 m/s[/tex]

After collision 2 kg mass deviate 40 degree from x axis

Let [tex]v_1=1.9[/tex] m/s be the velocity of 2 kg piece and [tex]v_2[/tex] be the of second Piece  after collision

Conserving momentum in x direction

[tex]m_1\times u_1+0=m_1v_1\cos \theta +mv_2\cos \theta _2[/tex]

[tex]4\times 2+0=2\times 1.9\cos 40+3v_2\cos \theta _2[/tex]

[tex]v_2\cos \theta_2=1.696[/tex]------1

Conserving Momentum in Y direction

[tex]m_1\times u_1+m_2\times u_2=m_1v_1\sin \theta -m_2v_2\sin \theta _2 [/tex]

[tex]0+0=2\times 1.9\sin 40-3\times v_2\sin \theta _2[/tex]

[tex]v_2\sin \theta _2=0.8142[/tex]-------2  

squaring and then adding 1 & 2 we get

[tex](1.696)^2+(0.8142)^2=(v_2)^2\cdot (\cos^2 \theta _2+\sin ^2\theta _2)[/tex]

[tex]v=1.88 m/s[/tex]

for direction divide 1 and 2 we get

[tex]\tan \theta _2=0.48[/tex]

[tex]\theta _2=25.64^{\circ}[/tex]  

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