an alpha particle m=4 collides elastically with a calcium atomm=20 at rest. the

Question

an alpha particle m=4 collides elastically with a calcium atomm=20 at rest. the initial velocity of the alpha particle isVi m/s. after collision the alpha particle has a velocity10i-5j m/s. if the velocity of the calcium atom aftercollision is vxi+vyj find all the unknowns in the problem find the smallest coefficient of static friction between thetwo blocks to keep the 2 kg from falling. there is notfriciton between 4 kg block and table. -> 24 n |4 kg| |2kg| an alpha particle m=4 collides elastically with a calcium atomm=20 at rest. the initial velocity of the alpha particle isVi m/s. after collision the alpha particle has a velocity10i-5j m/s. if the velocity of the calcium atom aftercollision is vxi+vyj find all the unknowns in the problem find the smallest coefficient of static friction between thetwo blocks to keep the 2 kg from falling. there is notfriciton between 4 kg block and table. -> 24 n |4 kg| |2kg|

Explanation / Answer

as the collision is elastic the coefficient of restitution is 1 hence velocity of approach=velocity of separation hence vi=(vx-10)i+(vy+5)j hence vy=-5m/sec and v=vx-10 and conserving the momentum in the x direction we have 4v=4*10+20*Vx v=10+5vx solving both the equation we have everything

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