A block of mass m=7.6kg, moving on a frictionless surface with speed vi= 7.6m/s, makes a perfectly elastic collision witha block of mass M at rest. After the collision, the 7.6 kg block recoils with a speed of vf=2.5m/s. In the figure, the blocks are in contact for 0.20 s. The average force on the 7.6 kg block, while the two blocks are in contact, is the closest to: 241N, 194N, 289N, 336N, 384N
Keep it simple ..?
F x t = change of momentum (?mv)
For 7.6kg block ..
?(mv) = 7.60kg x (7.6m/s + 2.5m/s) = 76.76 kg.m/s .. (accounting
for the change in direction)
F x t = 76.76 kg.m/s
F = 76.76 / 0.20s .. .. .. ?F = 384 N .. (Ans. A)
force on 7.6kg block = change in momentum of 7.6 kb block / time = 7.6 * ( 7.6 + 2.5) / 0.2
= 383.8 N
change in mometum = 7.6( 2.5+7.6) = 76.676 = force *0.2
force = 383.8 N ~ 384 N
Lets say the 7.6 kg block has momentum is right direction.
Hence its initial momentum, mi = 7.6 * 7.6
It recoils after collision. so it has momentum finally in left direction.
Hence its final momentum, mf = -(7.6 * 2.5)
Now total change in momentum (impulse) = mi - mf
= 7.6 * 10.1 = 76.76
We know, Impulse = Average Force * Time
Hence, Average Force = Impulse/Time
= 76.76/0.20
Therefore, Average Force = 383.8 N
Which is close to 384 Newton.
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