Two identical carts-a and b-initially are moving on frictionless air tracks. the initial speed of cart a is twice that of cart b. you then exert the same constant force on the two carts over 1 second. one second later, the change in momentum of cart a is a) non-zero and twice the change in momentum of cart b b) non-zero and the same as the change m momentum of cart b c) zero d) non-zero and half the change in momentum of cart b e) not enough information to determine

b) non-zero and the same as the change m momentum of cart B

Explanation:

As we know that change in momentum is also known as impulse which is equal to the product of force and time interval

so we will have

since we applied same force on two carts for small time interval of t = 1 s

so the impulse is given as

so both the carts will have same change in momentum as we know that force and time interval is same for both

so correct answer is given as

b) non-zero and the same as the change m momentum of cart B

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm