Question

3. A 98.0 kg hockey player skating at 2.12 m/s 32° S of W is body checked by an 87.0 kg opponent who is skating at 1.40 m/s 42° N of E. They collide inelastically and stick together. a. What is the velocity of the pair immediately after the collision? Solve using both methods. .

Trig. Method (include a vector diagram)

Component Method (show all work)

b. Determine the impulse imparted to the 98.0 kg player. Show a complete vector diagram and label your answer.

c. From what you know about Impulse, what would the Impulse for the 87.0 kg player be? Provide both magnitude and direction. (you need not calculate if you understand the concepts)

d. State what you found out for questions b and c above and justify your results using I = Favg t. State any of the appropriate Newton’s Laws to help explain. No calculations are necessary. Simply justify what you found out from (b) and (c) by using the other equation for Impulse, I = Favg t.

Answer 1

a)

m₁ =98.0kg

v_1i = (2.12 m/s)

θ₁= 32 deg S of W = 212^o

v_1i= v_1ix + v_1iy =v_1i (cosθ₁+ sinθ₁)

v_1i =2.12 m/s*(cos(212) i + sin(212) j)

v_1i= (-1.80i - 1.12j) m/s

m₂ = 87.0 kg

v_2i = (1.40 m/s)

θ₂= 42 deg N of E = 42^o

v_2i= v_2ix + v_2iy =v_2i (cosθ₃+ sinθ₂)

v_1i =1.40m/s*(cos(42) i + sin(42) j)

v_1i= (1.04i +0.94j) m/s

Now use conservation of Momentum

m₁*v_1i + m₂*v_2i = m₁*v_1f + m₂*v_2f
Now when they stick together, v_1f = v_2f = v_common
m₁*v_1i + m₂*v_2i = v_f*(m₁+m₂)
v_f = (m₁*v_1i + m₂*v_2i) /(m₁+m₂)
Plug values,
v_common= (98.0 kg*(-1.80i - 1.12j m/s) + 87.0 kg*(1.04i + 0.94j m/s))/ (98.0 kg + 87.0 kg)
v_common = (-0.463i - 0.155j) m/s

b)

Impulse 1 = I₁= m₁*(v_common - v_1i)
I₁ = 98.0 kg*((-0.463i - 0.155j) - (-1.80i - 1.12j)) m/s
I₁ = (131i + 94.9j) kg.m/s

Magnitude I₁
|I₁| = sqrt(i^2 + j^2) = sqrt(131^2 + 94.9^2) kg.m/s = 162 kg.m/s

c)

I₂ = - I₁ = (-131i - 94.9j) kg.m/s
Magnitude I₂
|I₁| = sqrt(i^2 + j^2) = sqrt(131^2 + 94.9^2) kg.m/s = 162 kg.m/s

Direction of I₂

|θ_I2| = tan^-1(j/i)= tan^-1(-94.9/-131) = 35.92^o above +X axis

d)

We got relation,

I₂ = - I₁

F_ave2*t = -F_ave1*t

This is nothing but Newton’s 3^rd law of motion.

Which states that, every action force has a reaction force, both have equal magnitudes but directions are exactly opposite.