A 25-kg child runs at 4.0 m/s and jumps onto a shopping cart and holds on for dear life. The cart has mass 15 kg. Assuming the child rides on the cart after the collision, the speed of the child and shopping cart just after the child jumps on is:. a) zero
b) 2.0 m/s
c) 3.0 m/s
d) 2.5 m/s

Linear momentum (mass x speed) has to be conserved.

-- Momentum before the jump:

(boy's mass) x (boy's speed) = (25 kg) x (4.0 m/s) = 100 kg-m/s

(cart's mass) x (cart's speed) = (15 kg) x (zero) = zero

Total momentum before the jump:  (100 kg-m/s) + (zero) = (100 kg-m/s)

-- Momentum after the jump:

(mass of boy+cart) x (speed of boy+cart) = (40 kg) x (speed)

-- Momentum after the jump = momentum before the jump

(40 kg) x (speed) = 100 kg-m/s

Divide each side by  40 kg:

Speed = (100 kg-m/s) / (40 kg)

Speed = 2.5 m/s  (d)

unbalanced force

explanation:

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft