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Standard XIII

Physics

Block on a Block Problems

Block B of ma...

Question

Block B of mass $2 kg$ is placed on smooth horizontal plane. Block A of mass $1 kg$ is placed on block B. The coefficient of friction between A and B is 0.40. The block A is imparted a velocity $16 m/s$ at t = 0. Find the time at which momentum of the two blocks are equal (in seconds). $(g = 10 {m/s}^{2})$ .

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Solution

$P_{1} = m_{1} V_{1} = (1) (16 - μ g t)$ till velocities are equal.
$P_{2} = m_{2} V_{2} = (2) (\frac{μ m_{1} g}{m_{2}}) t = μ (1) g t$ till velocities are equal.
For $P_{1} = P_{2}, 16 - μ g t = μ g t$
$t = \frac{16}{(2) (0.4) (10)} = 2 sec$

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Block B of mass 2 kg is placed on smooth horizontal plane. Block A of mass 1 kg is placed on block B. The coefficient of friction between A and B is 0.40. The block A is imparted a velocity 16 m/s at t = 0. Find the time at which momentum of the two blocks are equal (in seconds). (g=10 m/s2).

P1=m1V1=(1)(16−μgt) till velocities are equal. P2=m2V2=(2)(μm1gm2)t=μ(1)gt till velocities are equal. For P1=P2,16−μgt=μgt t=16(2)(0.4)(10)=2 sec

$P_{1} = m_{1} V_{1} = (1) (16 - μ g t)$ till velocities are equal.
$P_{2} = m_{2} V_{2} = (2) (\frac{μ m_{1} g}{m_{2}}) t = μ (1) g t$ till velocities are equal.
For $P_{1} = P_{2}, 16 - μ g t = μ g t$
$t = \frac{16}{(2) (0.4) (10)} = 2 sec$