Answer:
s₁ = 0.022 m
Explanation:
From the law of conservation of momentum:
![m_1u_1 + m_2u_2 = m_1v_1+m_2v_2](https://tex.z-dn.net/?f=m_1u_1%20%2B%20m_2u_2%20%3D%20m_1v_1%2Bm_2v_2)
where,
m₁ = mass of hockey player = 97 kg
m₂ = mass of puck = 0.15 kg
u₁ = u₂ = initial velocities of puck and player = 0 m/s
v₁ = velocity of player after collision = ?
v₂ = velocity of puck after hitting = 48 m/s
Therefore,
![(97\ kg)(0\ m/s)+(0.15\ kg)(0\ m/s)=(97\ kg)(v_1)+(0.15\ kg)(48\ m/s)\\\\v_1 = -\frac{(0.15\ kg)(48\ m/s)}{97\ kg} \\v_1 = - 0.074 m/s](https://tex.z-dn.net/?f=%2897%5C%20kg%29%280%5C%20m%2Fs%29%2B%280.15%5C%20kg%29%280%5C%20m%2Fs%29%3D%2897%5C%20kg%29%28v_1%29%2B%280.15%5C%20kg%29%2848%5C%20m%2Fs%29%5C%5C%5C%5Cv_1%20%3D%20-%5Cfrac%7B%280.15%5C%20kg%29%2848%5C%20m%2Fs%29%7D%7B97%5C%20kg%7D%20%5C%5Cv_1%20%3D%20-%200.074%20m%2Fs)
negative sign here shows the opposite direction.
Now, we calculate the time taken by puck to move 14.5 m:
![s_2 =v_2t\\\\t = \frac{s_2}{v_2} = \frac{14.5\ m}{48\ m/s} \\\\t = 0.3\ s](https://tex.z-dn.net/?f=s_2%20%3Dv_2t%5C%5C%5C%5Ct%20%3D%20%5Cfrac%7Bs_2%7D%7Bv_2%7D%20%3D%20%5Cfrac%7B14.5%5C%20m%7D%7B48%5C%20m%2Fs%7D%20%5C%5C%5C%5Ct%20%3D%20%200.3%5C%20s)
Now, the distance covered by the player in this time will be:
![s_1 = v_1t\\s_1 = (0.074\ m/s)(0.3\ s)](https://tex.z-dn.net/?f=s_1%20%3D%20v_1t%5C%5Cs_1%20%3D%20%280.074%5C%20m%2Fs%29%280.3%5C%20s%29)
<u>s₁ = 0.022 m</u>
Answer:
a. Δx = 2.59 cm
Explanation:
mb = 0.454 kg , mp = 5.9 x 10 ⁻² kg , vp = 8.97 m / s , k = 21.0 N / m
Using momentum conserved
mb * (0) + mp * vp = ( mb + mp ) * vf
vf = ( mp / mp + mb) * vp
¹/₂ * ( mp + mb) * (mp / mp +mb) ² * vp ² = ¹/₂ * k * Δx²
Solve to Δx '
Δx = √ ( mp² * vp² ) / ( k * ( mp + mb )
Δx = √ ( ( 5.9 x 10⁻² kg ) ² * (8.97 m /s) ² / [ 21.0 N / m * ( 5.9 x10 ⁻² kg + 0.454 kg ) ]
Δx = 0.02599 m ⇒ 2.59 cm
Answer:
i wanna say tranverse if not surface
Explanation: