Question

You and your friends are doing physics experiments on a frozen pond that serves as a frictionless, horizontal surface. Sam, with mass 78.0 kg , is given a push and slides eastward. Abigail, with mass 58.0 kg , is sent sliding northward. They collide, and after the collision Sam is moving at 40.0 ∘ north of east with a speed of 6.20 m/s and Abigail is moving at 21.0 ∘ south of east with a speed of 8.20 m/s .

(a) What was the speed of each person before the collision?
(b) By how much did the total kinetic energy of the two people decrease during the collision?

Answer 1

(a) the speed of Sam was 10.44 m/s and

the speed of Abigail was 2.42 m/s

(b) Decreased by 973.158 J or ΔKE = -973.158 J

Step-by-step explanation:

Here we have

Initial total momentum = final total momentum

Final total momentum = m₁ 6.2(sin 40 i + cos 40 j) + m₂ 8.2 (-sin 21 i + cos 21 j)

Where m₁ = Mass of Sam = 78.0 kg

m₂ = Mass of Abigail = 58.0 kg

Therefore

m₁ 6.2(sin 40 i + cos 40 j) + m₂ 8.2 (-sin 21 i + cos 21 j)

310.85 i 370.459 j + -170.44 i + 444.01 j

= 140.41 i + 814.5 j

(a) Therefore since the direction of motion of Sam is in the x direction and Abigail in the y direction, we have

we have

78 × v₁ₓ + 58 × 0 = 78 × 4.75 + 58×7.66

v₁ₓ = 10.44 m/s

Similarly

78×0 + 58×v₂ₙ = 78 × 3.99 - 58× 2.94

v₂ₙ = 140.4/58 = 2.42 m/s

Therefore the speed of Sam was 10.44 m/s and

the speed of Abigail was 2.42 m/s

Their total kinetic energy before collision was

KE initial = 0.5×78×10.44² + 0.5×58×2.42²

= 4422.28 J

Their Kinetic energy after collision is

0.5×78×6.2² + 0.5×58×8.2² = 3449.12

Therefore, their Kinetic energy decreased by

4422.28 J - 3449.12 J = 973.158 J

Answer 2

a) v1=6.17m/s v2=1.26m/s

b) 14.5J

Step-by-step explanation:

a) In this case we have to take into account the momentum conservation and energy conservation.

`p_i=p_f\\\\m_1v_1+m_2v_2=m1v_1'+m_2v_2'\\\\`

Furthermore, the x component and vertical component of the total momentum has to conserve:

`p_(ix)=p_(fx)\\\\m_1v_1+0=m_1v_1'cos\theta_1+m_2v_2'cos\theta_2\\\\p_(iy)=p_(fy)\\\\m_2v_2=m_1v_1'sin\theta_1-m_2v_2'sin\theta_2`

By replacing the values of m1, m2, v1' and v2' in both equations for px and py we obtain (it is necessary to take into account the correct angle theta1 and theta2 are angles respect to the horizontal axis.):

theta1 = 90-40=50°

theta2 = 90-21=69°

`(78kg)v_1=(78kg)(6.20m/s)cos50\°+(58kg)(8.2m/s)cos69\°\\\\v_1=6.17m/s`

and for v2:

`(58kg)v_2=(78kg)(6.2m/s)sin50\°-(58kg)(8.2m/s)sin69\°\\\\v_2=-1.26m/s`

the minus sing is because of the direction of v2.

b) The losses in energy can be computed by using the expression:

`\Delta E_k=E_(ik)-E_(fk)`

Hence, by replacing we get:

`E_(ik)=(1)/(2)(m_1v_1^2+m_2v_2^2)=1530.7J\\\\E_(fk)=(1)/(2)(m_1v_1'^2+m_2v_2'^2)=1545.2J\\\\\Delta E_(k)=14.5J`

hope this helps!