Question

A.149 kg baseball moving at 17.7 m/s is caught by a 57 kg catcher at rest on an ice skating rink, wearing

frictionless skates. With what speed does the catcher slide on the ice?

Do NOT put in units or it will be marked wrong! The answer's value only! Please round each answer to 3 places.

MaVa + MbVb = (Ma+b)(Va+b)

Answer 1

Answer:

v = 12.8 m/s

Explanation:

Applying the law of conservation of energy:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

m₁ = mass of baseball = 149 kg

m₂ = mass of catcher = 57 kg

u₁ = initial speed of ball = 17.7 m/s

u₂ = initial speed of catcher = 0 m/s

v₁ = v₂ = v = final speed of ball and the final speed of catcher = ?

both are same because ball is in hands of cathcer in the final state.

Therefore,

$(149\ kg)(17.7\ m/s)+(57\ kg)(0\ m/s)=(149\ kg)(v)+(57\ kg)(v)\\\\v = \frac{2637.3\ kgm/s}{206\ kg}$

v = 12.8 m/s