Chapter 11: Q. 40 (page 289)

A 500 g cart is released from rest 1.00 m from the bottom of a frictionless, 30.0° ramp. The cart rolls down the ramp and bounces off a rubber block at the bottom. FIGURE P11.40 shows the force during the collision. After the cart bounces, how far does it roll back up the ramp?

Expert verified

The distance where cart rolls back up to the ramp is $49.8 c m$

Step by step solution

Mass of cart, $m = 500 g$

A plot of force v/s time.

Impulse in the given case would be the area under the curve,

$J_{x} = \frac{1}{2} \times b a s e \times h e i g h t J_{x} = \frac{1}{2} \times 200 \times 26.7 \times 10^{- 3} = 2.67 N . s$

According to the equation of motion,

$v_{i} = \sqrt{2 g x_{i} \sin θ} v_{i} = \sqrt{2 \times 9.8 \times 1 \times \sin 30} v_{i} = 3.13 m / s$

Similarly, the final velocity is,

$v_{f} = \sqrt{2 g x_{f} \sin θ} v_{f} = \sqrt{2 \times 9.8 \times x_{f} \times \sin 30} v_{f} = 3.13 \sqrt{x_{f}}$

As we know that, $J_{x} = △ p$

Hence,

$x_{f} = {(\frac{2.67 - ((0.5) \times (- 3.13))}{3.13 \times 0.5})}^{2} x_{f} = 49.8 c m$

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