1)Instead of hitting a baseball, a baseball player hits a very flexible, rubber

Question

1)Instead of hitting a baseball, a baseball player hits a very flexible, rubber baseball

(m = 0.15 kg)

similar to the situation shown in the figure below. The ball is initially traveling horizontally with a speed of 40 m/s. The batter hits a fly ball as shown, with speed

vf = 61 m/s.

(b) If the contact time changes (increases) by a factor of 30 but the initial and final velocities are the same, by what factor does the force of the bat on the ball change? (F1 is the force with the original contact time and F2 is the force with the increased contact time.)

=

2)Consider the collision between two hockey pucks in the figure below. They do not stick together. Their speeds before the collision are v1i = 30 m/s and v2i = 13 m/s. It is found that after the collision one of the pucks is moving along x with a speed of 7 m/s. What is the final velocity of the other puck?

Explanation / Answer

Part b)

This problem can be solved with expressions of momentum and impulse

I = F dt

we can consider that F = constant during the contact

I = F t

also

I = p

and tell us that speed remains the same is the two cases, this implies that I = cte

I1 = F1 t

I2 = F2 t2

t2 = 30 t1

I2 = F2 30 t1 = 30 F2 t1

I1 =I2

F1 t1 = 30 F2 t1

F1/F2 = 30

F2/F1 = 1/30

Result The force decreases factor 30

Part c)

This problem is to be solved with the conservation of momentum elastic collision

before the crash

P1i = m v1i bold letters indicate vectors

decompose on each axis

Cos 30 = V1ix/V1 V1ix = V1 cos 30

Sin 30 = V1iy/v1 V1iy = V1 Sin 30

P1x = m V1 Cos 30

P1y = - m V1 Sin 30

P2 = m V2

P2y = m V2y

P2x =0

After the crash

P1fx = m V1fx

P2fx = m V2fx

P2fy = m V2fy

We write conservation of momentum on each axis

Axis X

Pix = Pfx

m V1 Cos 30 + 0 = m V1fx +m V2fx

V1i Cos 30 = V1fx + V2fx

V1i Cos 30 - V1fx = V2fx

Data

V1 = 30 m/s

V2y =13 m/s

V1fx = 7 /s

m1 = m.2 = cte

V2fx = 30 Cos 30 – 7

V2fx = 18.98 m/s

Axis Y

Piy = Pfy

- m V1 Sin 30 + m V2iy = 0 + m V2fy

- 30 Sin 30 + 13 = V2fy

V2fy = -2 m/s

We can give the answer in two ways

V2f = (18.98 i – 2 j ) m/s

or they ask module and angle, using Pythagoras and trigonometry

V2f2= V2fx2+ V2fy2

V2f2 = 18.982 + (-2)2 = 364.24

V2f = 19.09 m/s

Tan = V2fy/V2fx

Tan = -2/18.98 = -0.105

= 6 deg

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.