A 40 gram bullet is fired with a velocity of 480 ms into blo

A 40 gram bullet is fired with a velocity of 480 m/s into block A, which has a mass of 4.5 kg. The coefficient of kinetic friction between block A and the cart is 0.45; there is no friction between the cart wheels and the ground. Knowing that the cart is initially at rest, has a mass of 6.0 kg, and can roll freely, determine: the final velocity of the cart and block. the final position of the block on the cart 3

Solution

When the body A is in motion the frictional force acts the body backwards i.e., that frictional force pulls the cart forward.

The frictional force acted by the body A = Coefficient of friction*Normal reaction acting on the body.

Normal reaction acting on the body by the cart is equals to the weight of the body.

Weight of the body= 4.5* 9.81=44.145N

The frictional force acted by the body= (0.45*44.145)=19.86525N

Given that the mass of the cart=6Kg.

Combined mass of the cart and block =10.5Kg

Force is defined as the rate of change in momentum i.e., The forve acting on the cart is the rate of change in its momemtum.

i.e., 10.5(v-u)=19.86525;

v is the final velocity of cart and blockA & u is the intial velocityof the cart and blockA;

10.5(v-0)=19.86525;

v=1.8919m/s.

The final velocity of the cart and the block is 1.8191m/s.

Law of conservation of Momemtum:m₁u₁+m₂v₁=m₂u₂+m₂v₂

(0.04*480)+0=(4.54)v₂;

v₂=4.22m/s.

Work against friction = force of friction x distance = 19.86525*x;

Kinetic energy of the block= 0.5*4.54*4.22² =40.56J

The Kinetic energy of the block should be equal to the work against the friction.

19.86525*x=40.56;

x=2.043m

Therefore the distance travellled by the bock before it comes to the rest is 2.04 meters.