On a frictionlesssurface, a 0.35 kg puck moves horizontally to the right (at ana

Question

On a frictionlesssurface, a 0.35 kg puck moves horizontally to the right (at anangle of 0°) and a speed of2.3 m/s. It collides with a 0.23 kg puck that isstationary. After the collision, the puck that was initially movinghas a speed of 2.0 m/s and is moving at an angle of32°. What is the velocity of the other puck after thecollision and the degrees?

So I know this
m1=.35kg
vi=2.3m/s
vf=2m/s

m2=.23kg
vi2=0m/s
vf2=?

How would you go about doing this without knowingacceleration?

Explanation / Answer

the conservation of linear momentum: in x-component : m1.vi= m1.vf.cos(-32) + m2.vf2.cos =>m2.vf2.cos = 0,805-0,7.cos(-32) y-component : 0=m1.vf.sin(-32) + m2.vf2.sin => m2.vf2.sin =-0,7.sin(-32) => (m2.vf2.sin)^2+(m2.vf2.cos )^2 = (m2.vf2)^2= (0,805-0,7.cos(-32))^2 +( -0,7.sin(-32)^2 =0,1823 => vf2=1,856 m/s => sin=-0,7.sin(-32)/(m2.vf2) =0,8689 =>=60,34o

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