Question

A novelty golf ball of mass m is launched with an initial velocity v0 = (25i + 13j) m/s and then follows a parabolic trajectory. At the top of the ball’s trajectory, it explodes into two fragments A and B. Fragment A has mass mA = 1/3 m and is stationary immediately after the explosion, while Fragment B has mass mB = 2/3 m and has non-zero velocity vB immediately after splitting from Fragment A. In answering the following questions, ignore air resistance and assume the terrain over which the ball and fragments fly is level.(a) How many seconds after the launch does the ball attain its maximum height? (b) What is the velocity v of the ball immediately prior to the explosion? (c) What is the velocity of Fragment B immediately before it hits the ground?

Answer 1

Explanation:

(a)

The initial vertical velocity is 13 m/s.  At the maximum height, the vertical velocity is 0 m/s.

v = at + v₀

0 = (-9.8) t + 13

t ≈ 1.33 s

(b)

Immediately prior to the explosion, the ball is at the maximum height.  Here, the vertical velocity is 0 m/s, and the horizontal velocity is constant at 25 m/s.

v = √(vx² + vy²)

v = √(25² + 0²)

v = 25 m/s

(c)

Momentum is conserved before and after the explosion.

In the x direction:

m vx = ma vax + mb vbx

m (25) = (⅓ m) (0) + (⅔ m) (vbx)

25m = (⅔ m) (vbx)

25 = ⅔ vbx

vbx = 37.5 m/s

And in the y direction:

m vy = ma vay + mb vby

m (0) = (⅓ m) (0) + (⅔ m) (vby)

0 = (⅔ m) (vby)

vby = 0 m/s

Since the vertical velocity hasn't changed, and since Fragment B lands at the same height it was launched from, it will have a vertical velocity equal in magnitude and opposite in direction as its initial velocity.

vy = -13 m/s

And the horizontal velocity will stay constant.

vx = 37.5 m/s

The velocity vector is (37.5 i - 13 j) m/s.  The magnitude is:

v = √(vx² + vy²)

v = √(37.5² + (-13)²)

v ≈ 39.7 m/s