An archer shoots an arrow toward a 300-g target that is sliding in her direction at a speed of 2.30 m/s on a smooth, slippery surface. The 22.5-g arrow is shot with a speed of 38.0 m/s and passes through the target, which is stopped by the impact. What is the speed of the arrow after passing through the target

4.79m/s

Explanation:

According to law of conservation of momentum;

The sum of momentum of the bodies before collision is equal to the momentum after collision.

m1u1 + m2u2 = (m1+m2)v

Given;

m1 = 0.3kg

u1 = 2.30m/s

m2 = 0.0225kg

u2 = 38m/s

Required

speed of the arrow after passing through the target v

Substituting the given data into the formula

0.3(2.3) + 0.0225(38) = (0.3 + 0.0225)v

0.69 + 0.855 = 0.3225v

1.545 = 0.3225v

v = 1.545/0.3225

v = 4.79m/s

Hence the speed of the arrow after passing through the target is 4.79m/s

the force required to move the cart is divided by three or 33.33 n.

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y