Question #168949

A 70.0-kg ice hockey goalie, originally at rest, catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/s. Suppose the goalie and the ice puck have an elastic collision and the puck is reflected back in the direction from which it came. What would their final velocities be in this case?


vpuck =

vgoalie =



1
Expert's answer
2021-03-08T08:29:14-0500

Let the mass of the hockey goalie be m1m_1 and the mass of the hockey puck be m2m_2. Let the initial speed of the hockey goalie be u1u_1 and the initial speed of the hockey puck be u2u_2. From the law of conservation of momentum, we have:


m1u1+m2u2=m1v1+m2v2.(1)m_1u_1+m_2u_2=m_1v_1+m_2v_2. (1)

Since collision is elastic, kinetic energy is conserved and we can write:


12m1u12+12m2u22=12m1v12+12m2v22.(2)\dfrac{1}{2}m_1u_1^2+\dfrac{1}{2}m_2u_2^2=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_2^2. (2)

Let’s rearrange equations (1) and (2):


m1(u1v1)=m2(v2u2),(3)m_1(u_1-v_1)=m_2(v_2-u_2), (3)m1(u12v12)=m2(v22u22).(4)m_1(u_1^2-v_1^2)=m_2(v_2^2-u_2^2). (4)

Let’s divide equation (4) by equation (3):


(u1v1)(u1+v1)u1v1=(v2u2)(v2+u2)v2u2,\dfrac{(u_1-v_1)(u_1+v_1)}{u_1-v_1}=\dfrac{(v_2-u_2)(v_2+u_2)}{v_2-u_2},u1+v1=u2+v2.(5)u_1+v_1=u_2+v_2. (5)

Let's express v2v_2 from the equation (5) in terms of u1u_1u2u_2 and v1v_1:


v2=u1u2+v1.(6)v_2=u_1-u_2+v_1. (6)

Let’s substitute equation (6) into equation (3). After simplification, we get:


(m1m2)u1+2m2u2=(m1+m2)v1.(m_1-m_2)u_1+2m_2u_2=(m_1+m_2)v_1.

From this equation we can find the final speed of the hockey goalie, v1v_1:


v1=(m1m2)(m1+m2)u1+2m2(m1+m2)u2.v_1=\dfrac{(m_1-m_2)}{(m_1+m_2)}u_1+\dfrac{2m_2}{(m_1+m_2)}u_2.


Since, u1=0u_1=0 (hockey goalie, originally at rest), we get:


v1=(m1m2)(m1+m2)u1+2m2(m1+m2)u2.v_1=\dfrac{(m_1-m_2)}{(m_1+m_2)}u_1+\dfrac{2m_2}{(m_1+m_2)}u_2.v1=20.150 kg(70 kg+0.150 kg)(35.0 ms)=0.150 ms.v_1=\dfrac{2\cdot0.150\ kg}{(70\ kg+0.150\ kg)}\cdot(-35.0\ \dfrac{m}{s})=-0.150\ \dfrac{m}{s}.

The sign minus means that the hockey goalie moves to the left after collision.

Substituting v1v_1into the equation (6) we can find the final speed of the hockey puck, v2v_2:


v2=2m1(m1+m2)u1+(m2m1)(m1+m2)u2,v_2=\dfrac{2m_1}{(m_1+m_2)}u_1+\dfrac{(m_2-m_1)}{(m_1+m_2)}u_2,v2=(m2m1)(m1+m2)u2,v_2=\dfrac{(m_2-m_1)}{(m_1+m_2)}u_2,v2=(0.150 kg70 kg)(0.150 kg+70 kg)(35 ms)=34.9 ms.v_2=\dfrac{(0.150\ kg-70\ kg)}{(0.150\ kg+70\ kg)}\cdot(-35\ \dfrac{m}{s})=34.9\ \dfrac{m}{s}.

The sign plus means that the hockey puck moves to the right after collision.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS