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Two small steel balls A and B have mass 0.6 kg and 0.2 kg respectively - Edexcel - A-Level Maths Mechanics - Question 2 - 2005 - Paper 1
Question 2
Two small steel balls A and B have mass 0.6 kg and 0.2 kg respectively. They are moving towards each other in opposite directions on a smooth horizontal table when t... show full transcript
Worked Solution & Example Answer:Two small steel balls A and B have mass 0.6 kg and 0.2 kg respectively - Edexcel - A-Level Maths Mechanics - Question 2 - 2005 - Paper 1
Step 1
the speed of A immediately after the collision
Answer
To find the speed of ball A after the collision, we can use the principle of conservation of momentum. The formula for momentum is:
Before the collision, the total momentum of the system can be calculated as:
= 4.8 ext{ kg m s}^{-1} - 0.4 ext{ kg m s}^{-1} = 4.4 ext{ kg m s}^{-1} $$ Let the speed of A after the collision be $v$ (m s⁻¹). Since the speed of B after the collision is twice that of A, it can be expressed as $2v$. Therefore, the total momentum after the collision can be expressed as: $$ p_{final} = (0.6 ext{ kg} imes v) + (0.2 ext{ kg} imes 2v) \ = 0.6v + 0.4v = 1.0v $$ Setting the two moments equal to each other (conservation of momentum): $$ 4.4 = 1.0v $$ Solving for $v$ gives: $$ v = 4.4 ext{ m s}^{-1} $$Step 2
the magnitude of the impulse exerted on B in the collision
Answer
Impulse is defined as the change in momentum of an object. The impulse exerted on B can be calculated as:
= (0.2 ext{ kg} imes 2v) - (0.2 ext{ kg} imes -2) $$ Substituting the value of $v$: $$ I_B = (0.2 ext{ kg} imes 2 imes 4.4) - (0.2 ext{ kg} imes -2) \ = (0.2 imes 8.8) + (0.4) \ = 1.76 + 0.4 = 2.16 ext{ Ns} $$ Thus, the magnitude of the impulse exerted on B is 2.16 Ns.