A red wagon with a mass of 7\, {\text{kg}}7kg7, start text, k, g, end text is moving with a velocity of 4\,\dfrac{\text{m}}{\text{s}}4 s m 4, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction straight toward a blue wagon at rest. The red wagon’s mass is 777 times the mass of the blue wagon. The final velocity of the red wagon is 3\,\dfrac{\text{m}}{\text{s}}3 s m 3, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction . What is the velocity of the blue wagon immediately after the red wagon collides with it?

a. 6.69m/s

b. y=4.48m

c. t=1.43secs

Explanation:

Data given, acceleration,a=35m/s^2

distance covered,d=64cm=0.64m,

a. to determine the speed, we use the equation of motion

initial velocity,u=0m/s

if we substitute values we arrive at

b. After taking the shot,the acceleration value is due to gravity i.e a=9.81m/s^2

and the distance becomes (y-2.2) above the ground. When it reaches the maximum height, the final velocity becomes zero and the initial velocity becomes 6.69m/s.

Hence we can write the equation above again

if we substitute values we have

c. the time it takes to arrive at 1.83m is obtain by using the equation below

if we insert the values, we solve for t , hence t=1.43secs