Question

Part Q: A block of mass m moving due east at speed v collides with and sticks to a block of mass 2m that is moving at the same speed v but in a direction 45.0∘ north of east. Find the direction in which the two blocks move after the collision.

Express your answer as angle theta in degrees measured north of east.

Answer 1

According to law of conservation of linear momentum, linear momentum of a system doesn't change if no net external force is applied on the system.

Also, linear momentum=mv_x i + mv_yj where m is mass, v_x and v_y are velocity components in x and y directions respectively.

Consider east as positive x and north as positive y.

Initial momentum=mv i + 2m*v*cos45 i + 2m*v*sin45 j =( mv + 2m*v*cos45)  i + 2m*v*sin45 j

Finally,the masses stick together and move with a common velocity V_xi + V_yj

So, final momentum=3m(V_xi + V_yj) =3mV_xi + 3mV_yj

Initial momentum=final momentum.

So, 3mV_xi + 3mV_yj=( mv + 2m*v*cos45)  i + 2m*v*sin45 j

So, 3mV_x= mv + 2m*v*cos45 and 3mV_y =  2m*v*sin45

=> V_x=(mv + 2m*v*cos45 )/(3m)=(v + 2v*cos45 )/3 and V_y=  2m*v*sin45 /(3m)= 2v*sin45 /3

So, required angle=tan^-1(V_y/V_x)=tan^-1{(2v*sin45 /3)/[(v + 2v*cos45 )/3]}=tan^-1{(2sin45 )/(1 + 2cos45 )}=30.36 deg.