1 A baseball player hits a baseball m 0149 kg as shown in t

1) A baseball player hits a baseball

(m = 0.149 kg) as shown in the figure below. The ball is initially traveling horizontally with speed of 38 m/s. The batter hits a fly ball as shown, with a speed v_f = 53 m/s.

(b) If the ball and the bat are in contact for a time of 8.0 ms, what is the magnitude of the average force of the bat on the ball?

Compare this answer to the weight of the ball.

(c) What is the impulse imparted to the bat

2) A golf ball is hit from the tee and travels a distance of 280 yards. Estimate the magnitude of the impulse imparted to the golf ball. Ignore air drag in your analysis. (Assume 1 yard = 0.9144 m. Assume the golf ball is hit at an angle that will yield maximum range.) Estimate the mass of a golf ball. (estimated mass= 56g)

Calculate the magnitude of the impulse. (Use your estimate.)

3) Two billiard balls undergo an elastic collision as shown in the figure below. Ball 1 is initially traveling along x with a speed of 13 m/s, and ball 2 is at rest. After the collision, ball 1 moves away with a speed of 6.5 m/s at an angle

? = 60°. (For the following questions, assume the mass of ball 1 is equal to the mass of ball 2.)

Find the speed of ball 2 after the collision ....?

What angle does the final velocity of ball 2 make with the x axis?

4) Consider the collision between two hockey pucks in the figure below. They do not stick together. Their speeds before the collision are v_1i = 21 m/s and v_2i = 12 m/s. It is found that after the collision one of the pucks is moving along x with a speed of 5 m/s. What is the final velocity of the other puck?

Magnitude ......m/s?

Direction..... counterclockwise from the +x-axis

5) Two particles of mass m₁ = 1.7 kg and m₂ = 3.0 kg undergo a one-dimensional head-on collision as shown in the figure below. Their initial velocities along x are v_1i = 11 m/s and v_2i = -7.4 m/s. The two particles stick together after the collision (a completely inelastic collision). (Assume to the right as the positive direction.)

(a) Find the velocity after the collision.
m/s

(b) How much kinetic energy is lost in the collision?
J

Solution

Q1)

a)

The change in momentum in the positive x direction is

0.149* {53 cos 45 – (- 38)) = 11.246 Ns

The change in momentum in the direction perpendicular to the initial direction of motion is
0.149* (53sin 45 – 0) = 5.584 Ns

Resultant of these two changes in momentum which is the impulse is

0.149* 11.246 ² + 5.584 ²) = 1.87085 Ns

Tan = 5.584 / 11.246

= 26.41°
========================
b)

force = change in momentum / time =1.87 / 8e-3 = 233.75 N
=====================
c)
- 1.87 Ns
-26.41° = 339.59°
==================================

Please post other questions separately.