Photo AI

A railway truck P of mass 2000 kg is moving along a straight horizontal track with speed 10 m s⁻¹ - Edexcel - A-Level Maths Mechanics - Question 1 - 2003 - Paper 1

Question icon

Question 1

A-railway-truck-P-of-mass-2000-kg-is-moving-along-a-straight-horizontal-track-with-speed-10-m-s⁻¹-Edexcel-A-Level Maths Mechanics-Question 1-2003-Paper 1.png

A railway truck P of mass 2000 kg is moving along a straight horizontal track with speed 10 m s⁻¹. The truck P collides with a truck Q of mass 3000 kg, which is at r... show full transcript

Worked Solution & Example Answer:A railway truck P of mass 2000 kg is moving along a straight horizontal track with speed 10 m s⁻¹ - Edexcel - A-Level Maths Mechanics - Question 1 - 2003 - Paper 1

Step 1

a) the speed of P immediately after the collision

96%

114 rated

Answer

To calculate the speed of truck P immediately after the collision, we use the principle of conservation of momentum, which states that the total momentum before the collision equals the total momentum after the collision.

The initial momentum of the system is: extInitialMomentum=mPimesvP=2000imes10=20000extkgms1 ext{Initial Momentum} = m_P imes v_P = 2000 imes 10 = 20000 ext{ kg m s}^{-1}

The momenta after the collision can be expressed as: extFinalMomentum=mPimesvP+mQimesvQ=2000vP+3000imes5 ext{Final Momentum} = m_P imes v_P' + m_Q imes v_Q = 2000v_P' + 3000 imes 5

Setting the initial momentum equal to the final momentum gives: 20000=2000vP+1500020000 = 2000v_P' + 15000

Rearranging this equation, we find: 2000vP=20000150002000v_P' = 20000 - 15000 2000vP=50002000v_P' = 5000 v_P' = rac{5000}{2000} = 2.5 ext{ m s}^{-1}

Thus, the speed of truck P immediately after the collision is 2.5 m s⁻¹.

Step 2

b) the magnitude of the impulse exerted by P on Q during the collision

99%

104 rated

Answer

The impulse exerted by P on Q can be calculated using the formula: I=mimesextchangeinvelocityI = m imes ext{change in velocity}

The change in velocity of truck Q is: extChangeinvelocityforQ=vQ0=5extms1 ext{Change in velocity for Q} = v_Q - 0 = 5 ext{ m s}^{-1}

Therefore, the impulse exerted by P on Q is: I=mQimesvQ=3000imes5=15000extNsI = m_Q imes v_Q = 3000 imes 5 = 15000 ext{ Ns}

Hence, the magnitude of the impulse exerted by P on Q during the collision is 15000 Ns.

Join the A-Level students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;