Final speed of a kitten and ball of string

Question

An enthusiastic kitten collides with a ball of string that is rolled towards it. The 0.5 kg kitten is travelling at 0.5 m s^-1 due east before the collision and the 0.6 kg ball of string was travelling at 0.7 m s^-1 due west. If the kitten grabs the string during the collision and does not let go, what is the final speed of the kitten+string?

Answer 1

First thing to do is to generate a 2 dimensional coordinate axis using the info from the question.

Let EAST be POSITIVE and WEST be NEGATIVE.

second is to identify the mass and velocities of the kitten and ball BEFORE the collision

mk = 0.5 kg vk1 = 0.5 m/s (because the kat is moving EAST we let its velocity be POSITIVE

mb = 0.6 kg vb1 = -0.7 m/s (because the ball is moving WEST the velocity must be NEGATIVE

third we identify the mass and velocity AFTER the collision. note that the masses will not change, but because the kitten will not let go of the ball, they will have one combined super mass sharing the same velocity. Therefore after the collision we have one mass M and on velocity V. this is known as a perfect INELASTIC COLLISION

M = mk + mb = 0.5kg + 0.6kg = 1.1kg V = ?

Using the law of Conservation of momentum we know that the initial momentum before the collision must equal the momentum after the collision. in math form that looks like

(mk)(vk1) + (mb)(vb1) = MV

using algebra to solve for the unknown variable V

[(mk)(vk1) + (mb)(vb1)]/M = V

plugging in our numbers

[(0.5kg)(0.5m/s) +(0.6kg)(-0.7m/s)]/(1.1kg) = V

[(0.25 kg*m/s) +(-.42kg*m/s)]/(1.1kg) = V

[-0.17kg*m/s]/(1.1kg) = V

-0.15m/s = V

Because our final velocity is NEGATIVE, the ball and kitten will move together in the WEST direction at a velocity of 0.15m/s

(= ^_^=)