Question

Two cans of spam with identical masses collide. Before the collision, the hickory-smoke flavor is moving to the left at 4 m/s, while the hot-and-spicy flavor is moving to the right at 2 m/s. After the collision, the hickory-smoke is moving to the left at 1.2 m/s. What is the velocity of the hot-and-spicy?

Answer 1

Answer:

Explanation:

LET THE HICKORY -SMOKE CAN BE 1

               HOT AND SPICY CAN BE 2

MASS OF 1 = MASS OF 2 = M

LET THE RIGHTWARD DIRECTION BE TAKEN AS POSITIVE AND THE LEFTWARD DIRECTION BE TAKEN AS NEGATIVE

INITIAL VELOCITY OF 1 , U₁ = - 4 M/S

INITIAL VELOCITY OF 2 , U₂= + 2 M/S

FINAL VELOCITY  OF 1 , V₁ = -1.2 M/S

FINAL VELOCITY OF 2 , V₂ = ?

APPLYING CONSERVATION OF MOMENTUM BEFORE AND AFTER COLLISION.

M₁U₁+M₂U₂=M₁V₁+M₂V₂

M*(-4)+M*(2) = M*(-1.2)+M V₂

-2 M= M (-1.2+V₂)

V₂= - 2+1.2

V₂ = - 0.8 M/S

THEREFORE VELOCITY OF HOT AND SPICY CAN IS 0.8 M/S TOWARDS THE LEFT.

HOPE THIS HELPS.