Question

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 1

Step-by-step explanation:

`m_1=2 kg`

velocity of first piece
`u_1=4 m/s`

mass of second piece
`m_2=3 kg`

Velocity of second piece
`u_2=0 m/s`

After collision 2 kg mass deviate 40 degree from x axis

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

Conserving momentum in x direction

`m_1* u_1+0=m_1v_1\cos \theta +mv_2\cos \theta _2`

`4* 2+0=2* 1.9\cos 40+3v_2\cos \theta _2`

`v_2\cos \theta_2=1.696`------1

Conserving Momentum in Y direction

`m_1* u_1+m_2* u_2=m_1v_1\sin \theta -m_2v_2\sin \theta _2`

`0+0=2* 1.9\sin 40-3* v_2\sin \theta _2`

`v_2\sin \theta _2=0.8142`-------2

squaring and then adding 1 & 2 we get

`(1.696)^2+(0.8142)^2=(v_2)^2\cdot (\cos^2 \theta _2+\sin ^2\theta _2)`

`v=1.88 m/s`

for direction divide 1 and 2 we get

`\tan \theta _2=0.48`

`\theta _2=25.64^(\circ)`